1 © 2009 Brooks/Cole - Cengage More About Chemical Equilibria Acid-Base & Precipitation Reactions...

1

copy 2009 BrooksCole - Cengage

More About Chemical Equilibria

Acid-Base amp Precipitation

Reactions Chapter 15 amp 16

2


Objectives ndash Chapter 15bull Define the Common Ion Effect (151)bull Define buffer and show how a buffer

controls pH of a solution (152 153)bull Identify and Evaluate titration curves

(154)bull Use of Indicators (155)

3


Stomach Acidity ampAcid-Base Reactions

4


Acid-Base Reactions

bull Strong acid + strong base HCl + NaOH bull Strong acid + weak base

HCl + NH3 bull Weak acid + strong base HOAc + NaOH bull Weak acid + weak base

HOAc + NH3

What is relative pH

before during amp

after reaction

Need to study

a) Common ion

effect and buffers

b) Titrations

5


QUESTION What is the effect on the pH of adding NH4Cl to 025 M NH3(aq)

NH3(aq) + H2O NH4+(aq) + OH-(aq)

Here we are adding NH4+ an ion COMMON to the

equilibrium

Le Chatelier predicts that the equilibrium will shift to the left (1) right (2) no change (3)

The pH will go up (1) down (2) no change (3)

NH4+ is an acid

The Common Ion Effect

Section 151

6


Let us first calculate the pH of a 025 M NH3 solution

[NH3] [NH4+] [OH-]

initial 025 0 0change -x +x +xequilib 025 - x x x



pH of Aqueous NH3

7




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x2

025 - x

pH of Aqueous NH3

Assuming x is ltlt 025 we have

[OH-] = x = [Kb(025)]12 = 00021 M

This gives pOH = 267

and so pH = 1400 - 267

= 1133 for 025 M NH3

8


Problem What is the pH of a solution with 010 M NH4Cl and 025 M NH3(aq)


We expect that the pH will decline on adding NH4Cl Letrsquos test that

[NH3] [NH4+] [OH-]

initialchangeequilib

pH of NH3NH4+ Mixture

9





[NH3] [NH4+] [OH-]

initial 025 010 0change -x +x +xequilib 025 - x 010 + x x


10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x


Assuming x is very small

[OH-] = x = (025 010)(Kb) = 45 x 10-5 M

This gives pOH = 435 and pH = 965pH drops from 1133 to 965

on adding a common ion

11


Common Ion Effectbull The addition of a common ion in a base such as ammonia

limits the ionization of the base and the production of [OH-] (Le Chacirctelierrsquos Principle)

B(aq) + H2O(l) HB+(aq) + OH-(aq)bull For acid addition of conjugate base from salt limit

ionization of acid which limits [H3O+]

HA(aq) + H2O(l) A-(aq) + H3O+(aq)bull Doesnrsquot work if the conjugate base is from a strong acid

because the conjugate base is too weakbull In other words adding Cl- to above does not change the pH

because the Cl- comes from strong acid HCl which has a very large Ka

HCl + H2O H3O+ + Cl-

12


Common Ion Effect Practice

Calculate the pH of 030 M formic acid (HCO2H) in water and with enough sodium formate NaHCO2 to make the solution 010 M in the salt The Ka of formic acid is 18x10-4

HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+

13





With salt

14


Buffered SolutionsSection 152

HCl is added to pure water

HCl is added to a solution of a weak acid H2PO4

- and its conjugate base HPO4

2-

15


A buffer solution is a special case of the common ion effect

The function of a buffer is to resist changes in the pH of a solution either by H+ or by OH-

Buffer CompositionWeak Acid + Conj BaseHOAc + OAc-

H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16


Consider HOAcOAc- to see how buffers workACID USES UP ADDED OH-

We know that

OAc- + H2O HOAc + OH-

has Kb = 56 x 10-10

Therefore the reverse reaction of the WEAK ACID with added OH-

has Kreverse = 1 Kb = 18 x 109

Kreverse is VERY LARGE so HOAc completely consumes OH-

Buffer Solutions

17


Consider HOAcOAc- to see how buffers work

CONJ BASE USES UP ADDED H+

HOAc + H2O OAc- + H3O+

has Ka = 18 x 10-5

Therefore the reverse reaction of the WEAK BASE with added H+

has Kreverse = 1 Ka = 56 x 104

Kreverse is VERY LARGE so OAc- completely consumes H3O+

Buffer Solutions

18


Problem What is the pH of a buffer that has [HOAc] = 0700 M and [OAc-] = 0600 M


Ka = 18 x 10-5

Buffer Solutions

0700 M HOAc has pH = 245

The pH of the buffer will have

1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19


[HOAc] [OAc-] [H3O+]




Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20



equilib 0700 - x 0600 + x xAssuming that x ltlt 0700 and 0600 we have

Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21


Notice that the expression for calculating the H+ conc of the buffer is

[H3O+ ] = Orig conc of HOAc

Orig conc of OAc- x Ka

Buffer Solutions

Notice that the [H3O+] depends on

1the Ka of the acid2the ratio of the [acid] and

[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22


Henderson-Hasselbalch Equation

Take the negative log of both sides of this equation

The pH is determined largely by the pKa of the acid and then adjusted by the ratio of conjugate base and acid

or

Note HA and A- are generic terms for acid and conjugate base respectively

23


Henderson-Hasselbalch EquationWorks the same for base

B(aq) + H2O(l) HB+(aq) + OH-(aq)

Solve for [OH-] =

Take the negative log of both sides of this equation to get

The pOH (and pH) is determined largely by the pKb of the base and then adjusted by the ratio of the conjugate acid and base

119901119874119867=119901119870 b+119949119952119944 iquestiquest

24


Buffer Calculation Example(AcidConj Base)

What is the pH of a buffer that is 012 M in lactic acid HC3H5O3 and 010 M in sodium lactate (NaC3H5O3)

Ka for lactic acid is 14 10-4

25


Buffer Calculation Example

HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4

012 - xWersquoll make usual assumption that x is small

x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26


Buffer Calculation using the HendersonndashHasselbalch Equation

pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)

pH = 377 (same as before)

HC3H5O3 + H2O C3H5O3- + H3O+

acid conj base 012 010

27



pOH = ndashlog (18 x 10-5) + log(0045)(0080)

Calculate the pH of a buffer that has is 0080 M ammonia and 0045 M ammonium chlorideThe Kb for ammonia is 18 x 10-5 NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq) [base] [conj acid] 0080 0045

pOH = 474 + (ndash025)pOH = 449pH = 1400 ndash 449 = 951

28


Buffer Practice 1

Calculate the pH of a solution that is 050 M HC2H3O2 (acetic acid) and 025 M NaC2H3O2

(Ka = 18 x 10-5) Use H-H Eqn

29


Buffer Practice 2Calculate the concentration of sodium benzoate (NaC7H5O2) that must be present in a 020 M solution of benzoic acid (HC7H5O2) to produce a pH of 400

(Benzoic acid Ka = 63 x 10-5)

Use Henderson-Hasselbalch equation

HC7H5O2 + H2O H3O+ + C7H5O2-

30


Adding an Acid to a Buffer

(How a Buffer Maintains pH)Problem What is the pH when 100 mL of 100 M HCl

is added toa) 100 L of pure water (before HCl pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and

[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [HCl] after adding 100 mL of HCl to 100 L of

water

M1bullV1 = M2 bull V2

(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31



To play movie you must be in Slide Show Mode

32



Solution to Part (b)Step 1 mdash do the stoichiometry

H3O+ (from HCl) + OAc- (from buffer) rarr HOAc (from buffer)

The reaction occurs completely because K is very large

The stronger acid (H3O+) will react with the conj base of the weak acid (OAc-) to make the weak acid (HOAc)

What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (after HCl pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and

[OAc-] = 0600 M (pH before = 468)

33



Solution to Part (b) Step 1mdashStoichiometry

[H3O+] + [OAc-] [HOAc]

Before rxnChangeAfter rxnEquil []rsquos

What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 300)b) 100 L of buffer that has [HOAc] = 0700 M and

[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34



Solution to Part (b) Step 2mdashEquilibrium

HOAc + H2O H3O+ + OAc-

[HOAc] [H3O+] [OAc-]

Initial (M)Change (M)Equil (M) (re-established)


[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35






Equilibrium 0700-x x 0598+x

Because [H3O+] = 21 x 10-5 M BEFORE adding HCl we again neglect x relative to 0700 and 0598


[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)

pH = 474 -007 = 467The pH has not changed muchon adding HCl to the buffer

37


Adding a Base to a BufferProblem What is the pH when 100 mL of 100 M

NaOH is added toa) 100 L of pure water (before NaOH pH = 700)b) 100 L of buffer that has [HOAc] = 0700 M and

[OAc-] = 0600 M (pH = 468) (See slide 20)Solution to Part (a)Calc [OH-] after adding 100 mL of NaOH to 100 L of

water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38


Adding an Base to a Buffer

Solution to Part (b)Step 1 mdash do the stoichiometryOH- (from NaOH) + HOAc (from buffer) rarr

OAc- (from buffer and rxn)The reaction occurs completely because K is

very largeThe strong base (OH-) will react with weak acid

(HOAc) to make the conj weak base (OAc-)

What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (after NaOH pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and

[OAc-] = 0600 M (pH before = 468)

39


Adding a Base to a Buffer

Solution to Part (b) Step 1mdashStoichiometry [OH-] + [HOAc] rarr [OAc-]Before rxnChangeAfter rxnEquil []rsquos

What is the pH when 100 mL of 100 M NaOH is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]

= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40






Initial(M)Change (M)Equil (M)(re-established)


= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41







Because [H3O+] = 21 x 10-5 M BEFORE adding NaOH we again neglect x relative to 0600 and 0698

What is the pH when 100 mL of 100 M HCl is added to a) 100 L of pure water (pH = 1100)b) 100 L of buffer that has [HOAc] = 0700 M and [OAc-]

= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)

pH = 474 ndash 007 = 467The pH has not changed muchon adding NaOH to the buffer

43


Preparing a BufferYou want to buffer a solution at pH = 430

This means [H3O+] = 10-pH =10-430= 50 x 10-5 M

It is best to choose an acid such that [H3O+] is about equal to Ka (or pH asymp pKa)

mdashthen you get the exact [H3O+] by adjusting the ratio of acid to conjugate base

[H3O+ ] =

[Acid]

[Conj base] x Ka

44


You want to buffer a solution at pH = 430 or

[H3O+] = 50 x 10-5 M

POSSIBLE

ACIDSCONJ BASE PAIRS Ka

HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10

Best choice is acetic acid acetate

Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )

Therefore if you use 0100 mol of NaOAc and 0278 mol of HOAc you will have pH = 430

Solve for [HOAc][OAc-] ratio =

278

1

Preparing a Buffer

46


A final point mdashCONCENTRATION of the acid and conjugate

base are not as important as the RATIO OF THE NUMBER OF MOLES of

each

Result diluting a buffer solution does not change its pH

Preparing a Buffer

47


Commercial Buffers

bull The solid acid and conjugate base in the packet are mixed with water to give the specified pH

bull Note that the quantity of water does not affect the pH of the buffer

48


Buffer prepared from

84 g NaHCO3 (84g84 gmol)

weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49


Buffer ExamplePhosphate buffers are used a lot in labs to simulate blood pH of 740 As with any other buffer you need an acid and its conjugate base

In this case the lsquoacidrsquo is NaH2PO4 and the lsquobasersquo is Na2HPO4 So the equilibrium expression is

H2PO4-(aq) + H2O(l) HPO4

2-(aq) + H3O+(aq)

A) If the pKa of H2PO4- is 721 what should the

[HPO42-][H2PO4

-] ratio be

B) If you weigh 60 g of NaH2PO4 and make 500 mL of solution how much Na2HPO4 do you need add to make your pH 740 buffer

(assume the Na2HPO4 doesnrsquot affect the volume)

50


Buffer ExamplePart A


2-(aq) + H3O+(aq)

Can use H-H pH = pKa + log[HPO42-][H2PO4

-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-

=0155 molL 0500 L 142 g Na2HPO4mol

= 110 g Na2HPO4

51


Preparing a buffer PracticeUsing the acetic acid (molar mass 60 gmol) and sodium acetate (molar mass 82 gmol) buffer from before how much sodium acetate should you weigh to make 10 L of a pH 500 buffer Assume [HOAc] is 10 M in the final 10 L of buffer (pKa = 474)

52


Buffering CapacityChapter 153

bull The pH of a buffered solution is

determined by the ratio [A-][HA]bull As long as the ratio doesnrsquot change much

the pH wonrsquot change much

bull The more concentrated these two are the

more H+ and OH- the solution will be able to absorb

bull Larger concentrations rarr bigger buffer capacity

53


Buffer Capacity

bull Calculate the change in pH that occurs when 0010 mol of HCl is added to 10L of each of the following

bull 500 M HOAc and 500 M NaOAcbull 0050 M HOAc and 0050 M NaOAc

bull Ka= 18x10-5 (pKa = 474)bull Note Since [OAc-]=[HOAc] for both cases

the initial pH is 474

54


Large Buffer System

H3O+ + OAc- HOAc + H2O

H3O+ OAc- HOAc

Before Reaction (moles) 001 500 500

After Reaction (moles) 0 499 501

Stoichiometry Calculation

55


Calculate pH via Henderson-Hasselbalch Equations

At Equilibrium HOAc + H2O H3O+ + OAc-

0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738

Since pKa = 474 solution has gotten slightly more acidic

56


Small Buffer System

H3O+ + OAc- HOAc(aq) + H2O

H+ OAc- HOAc




57


Calculating pH for Small Buffer System

In table form (Equilibrium Re-established)

[HOAc] M [H3O+] M [OAcminus] M

Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x

Ignore xrsquos relative to 006 and 004 since they will be small Calculate via H-H eqn

[base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 004006]pH = 474 ndash 018 = 456

So solution is considerably more acidic than the pH 474 of the big buffer system

59


Mixing Acids and Bases(Stoichiometry amp

Equilibrium)Chapter 154

If you mix 500 mL of 010 M HCl and 500 mL of 010 NaOH what is the pHSince it is a strong acidstrong base what happens stoichiometrically

H+ + OH- H2O(l)

I(moles) 0005 0005 - After reaction 0 0 - Since you consume all acid and base and make water what is the pH 7 of course

60



Equilibrium)Now what if you mix 500 mL of 010 M HCl and 200 mL of 010 NaOH what is the pH

H+ + OH- H2O(l)

I(moles) 0005 0002 - After reaction (mol) 0003 0 -Total volume of solution is 700 mL (00700 L)So the [H+] = 0003 mol00700 L = 00429 M pH = -log[00429] = 137

61



Equilibrium)Now mix 500 mL of 010 M HCl and 700 mL of 010 NaOH what is the pH

H+ + OH- H2O(l)

I(moles) 0005 0007 - After reaction (mol) - 0002 -Total volume of solution is 1200 mL (01200 L)Since there was more OH- added than H+ all of the H+ is consumed and there is excess OH-[OH-] = 00020120 L = 0017 M OH-

pOH = 177 so pH = 1223

62



Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHSince it is a weak acidstrong base what happens stoichiometrically

HOAc + OH- H2O(l) + OAc-

I(moles) 0005 0005 - After reaction 0 0 - 0005Since HOAc is a weak acid and OH- is a strong base the OH- will still consume (take that H+) from the HOAc (The Stoichiometry part)

63



Equilibrium)If you mix 500 mL of 010 M HOAc and 500 mL of 010 NaOH what is the pHNow equilibrium will be re-established

H2O(l) + OAc- HOAc + OH-

Need Conc 0005 mol0100L I (conc) 005 0 0 C -x +x +x E 0050-x x xYou can think of this part as when we calculated pHrsquos of salts of weak acids It doesnrsquot matter how you get to this point the calculation is still the same

64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10

Usual assumption that x is small compared to 0050 so x = pOH = -log[527x10-6] = 528 so pH = 872

65





I(moles) 00050 00020 0 Reaction -00020 -00020 +00020After reaction 00030 0 00020

66



Equilibrium)If you mix 500 mL of 010 M HOAc and 200 mL of 010 NaOH what is the pHNow equilibrium will be re-established Since you have an acid and its conjugate base this is a buffer and you can use H-H eqn

H2O(l) + HOAc OAc- + H3O+

E 00030 00020 x pH = pKa + log [OAc-][HOAc] = 474 + log(0002)(0003) = 474 + log(0667) = 474 ndash 018 = 456

67


Titrations and pH CurvesTitrations and pH Curves

pHpH

Titrant volume mLTitrant volume mL

68


Titrations and pH Curves

Adding NaOH from the buret to acetic acid in the flask a weak acid In the beginning the pH increases very

slowly

69



Additional NaOH is added pH rises as equivalence point is approached

>

70



Additional NaOH is added pH increases and then levels off as NaOH is added beyond the equivalence

point

71


QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point


What is the pH at equivalence point


pH of solution of benzoic acid a weak acid

pH of solution of benzoic acid a weak acidBenzoic acid

+ NaOHBenzoic acid + NaOH

What is pH at half-way point


72



QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at the equivalence point

HBz + NaOH Na+ + Bz- + H2O

C6H5CO2H = HBz Benzoate ion = Bz-

73



The pH of the final solution will be1 Less than 72 Equal to 73 Greater than 7




74



The product of the titration of benzoic acid is the benzoate ion Bz-

Bz- is the conjugate base of a weak acid Therefore solution is basic at

equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75




pH at equivalence point is basic


Benzoic acid + NaOHBenzoic acid + NaOH

76



Strategy mdash find the conc of the conjugate base Bz- in the solution AFTER the titration then calculate pH

This is a two-step problem1stoichiometry of acid-base reaction2equilibrium calculation


77



STOICHIOMETRY PORTION1 Calc moles of NaOH reqrsquod(0100 L HBz)(0025 M) = 00025 mol HBzThis requires 00025 mol NaOH2Calc volume of NaOH reqrsquod00025 mol (1 L 0100 mol) = 0025 L25 mL of NaOH reqrsquod


78



STOICHIOMETRY PORTION25 mL of NaOH reqrsquod 3 Moles of Bz- produced = moles HBz =

00025 mol4 Calc conc of Bz-

There are 00025 mol of Bz- in a TOTAL SOLUTION VOLUME of


125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79



Equivalence PointMost important species in solution is benzoate

ion Bz- the weak conjugate base of benzoic acid HBz

Bz- + H2O HBz + OH- Kb = 16 x 10-10

[Bz-] [HBz] [OH-]initial 0020 0 0change- x +x +x equilib 0020 - x x x

QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at equivalence point

80



Equivalence PointMost important species in solution is benzoate ion Bz-

the weak conjugate base of benzoic acid HBz

Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81


QUESTION You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH to the equivalence point What is the pH at half-way point


pH at half-way point1 lt 72 = 73 gt 7


Equivalence point pH = 825


82




pH at half-way point

pH at half-way point Equivalence point

pH = 825


83



You titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH

What is the pH at the half-way point

[H3O+ ] = [HBz]

[Bz- ] x Ka

At the half-way point [HBz] = [Bz-]Therefore [H3O+] = Ka = 63 x 10-5

pH = 420 = pKa of the acid

Both HBz and Bz- are present

This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84


Titrations and pH CurvesYou titrate 100 mL of a 0025 M solution of benzoic acid with 0100 M NaOH

What was the pH when 60 mL of NaOH was added

StoichiometryBefore rxn moles 00006 00025 --After rxn mole 0 00019 00006After rxn [ ] 0 00019mol0106L 000060106 0 0018 00057 Equilbrium HBz + H2O H3O+ + Bz- pKa = 420EquilibriumNow can use H-H equationpH = 420 + log (000570018) = 420 ndash 50 = 370Remember this is essentially a buffer solution since HBz and Bz- are both in the solution

When doing these types of problems start from initial moles and go to the point you are considering

OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-



Practice What is the pH when 200 mL of NaOH are added

StoichiometryBefore rxn moles 00020 00025 --After rxn mole 0 00005 00020After rxn [ ] 0 00005mol0120L 000200120 0 00042 0017

Equilbrium HBz + H2O H2O + Bz- pKa = 420

EquilibriumNow can use H-H equationpH = 420 + log (001700042) = 420 + 061 = 481

86


Acetic acid titrated with NaOH

Weak acid titrated with a strong base

87


See Figure 184

Strong acid titrated with a strong base

88


Weak diprotic acid (H2C2O4) titrated with a strong base

(NaOH)

See Figure 186

89


Titration of a1 Strong acid with strong base2 Weak acid with strong base3 Strong base with weak acid4 Weak base with strong acid5 Weak base with weak acid6 Weak acid with weak base

pH

Volume of titrating reagent added --gt

90


Weak base (NH3) titrated with a strong

acid (HCl)

91


Acid-Base IndicatorsChapter 155

bull Weak acids that change color when they become bases

bull Since the Indicator is a weak acid it has a Ka

bull End point - when the indicator changes color

bull An indicator changes colors at pH=pKa1 pKa is the acid dissoc constant for the indicator

bull Want an indicator where pKa pH at equiv pt

92


Acid-Base IndicatorsAcid-Base Indicators

93


Indicators for Acid-Base Titrations


Since the pH change is large near the equivalence point you want an indicator that is one color before and one color after

94


Titration of a Base with an Acid

bull The pH at the equivalence point in these titrations is lt 7

bull Methyl red is the indicator of choice

95


Solubility and Complex Ion Equilibria

Chapter 16


Chapter 16

Lead(II) iodide

96


Objectives ndash Chapter 16

bull Define equilibrium constant for lsquoinsolublersquo salts (Ksp) (161)

bull Manipulate solubility by common ion effect (161)

bull Precipitation Reactions ndash comparing Q amp K of a reaction (162)

bull Define Complex ion amp show how complex ions affect solubility (163)

97


Types of Chemical Reactions

bull EXCHANGE REACTIONS AB + CD AD + CB

ndash Acid-base CH3CO2H + NaOH NaCH3CO2 + H2O

ndash Gas forming CaCO3 + 2 HCl CaCl2 + CO2(g) + H2O

ndash Precipitation Pb(NO3) 2 + 2 KI PbI2(s) + 2 KNO3

bull OXIDATION REDUCTION (Redox ndash Ch 18)ndash 4 Fe + 3 O2 2 Fe2O3

bull Apply equilibrium principles to acid-base and precipitation reactions

98


Analysis of Silver Group


All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver

GroupAlthough all salts formed in this experiment are

said to be insoluble they do dissolve to some

SLIGHT extent

AgCl(s) Ag+(aq) + Cl-(aq)

When equilibrium has been established no more

AgCl dissolves and the solution is SATURATED

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group


When solution is SATURATED expt shows that [Ag+] = 167 x 10-5 M

This is equivalent to the SOLUBILITY of AgCl

What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver

GroupAgCl(s) Ag+(aq) + Cl-(aq)

Saturated solution has

[Ag+] = [Cl-] = 167 x 10-5 M

Use this to calculate Kc

Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10

Because this is the product of ldquosolubilitiesrdquo we call it

Ksp = solubility product

constant

bull See Table 161 and Appendix A54bull Note The property lsquosolubilityrsquo is in molL

(Can also be expressed as gL or mgL)bull Solubility Product has no units

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103


Solubility ProductsIn general for the equilibrium for the reaction is expressed as

MmAa(s) mM+(aq) + aA-(aq)

The Ksp equilibrium expression is

Ksp =[M+]m[A-]a

The higher the Ksp the more soluble the salt is

104


Some Values of Ksp

105


Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq)

Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M

Solubility of Lead(II) Iodide


Consider PbI2 dissolving in water

PbI2(s) Pb2+(aq) + 2 I-(aq)

Calculate Ksp

if solubility = 000130 M

107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3



= 4 (solubility)3= 4 (solubility)3

Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat

3 The value we just calculated by

solubility gives a ball-park value

The actual Ksp of PbI2 is 98 x 10-9

Note Calculating Ksp from solubility

gives approximate values that can be off by a few OOMs due to ionic interactions



Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109


SolubilityKsp Practice 1bull Calculate the Ksp if the solubility of CaC2O4 is 61

mgL Note units here What should they be in order to determine Ksp (Molar Mass = 1281 gmol)

CaC2O4 (s) Ca2+(aq) + C2O42-(aq)

110


SolubilityKsp Practice 2bull Calculate the solubility of CaF2 in gL if the

Ksp is 39x10-11 (Molar Mass = 781 gmol)

111


The Common Ion EffectAdding an ion ldquocommonrdquo to an

equilibrium causes the equilibrium to shift back to reactant

112


Common Ion EffectPbCl2(s) Pb2+(aq) + 2 Cl-(aq)

Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10

(b) BaSO4 is opaque to x-rays Drinking a BaSO4 cocktail enables a physician to exam the intestines

(a) BaSO4 is a common mineral appearing a white powder or colorless crystals

114


Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2

Ksp for BaSO4 = 11 x 10-10

BaSO4(s) Ba2+(aq) + SO42-(aq)

Solution

Solubility in pure water = [Ba2+] = [SO42-] = x

Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M

Solubility in pure water = 11 x 10-5 molL


115


SolutionSolubility in pure water = 11 x 10-5 molL

Now dissolve BaSO4 in water already containing 0010 M Ba2+

Which way will the ldquocommon ionrdquo shift the equilibrium ___ Will solubility of BaSO4 be less than or greater than in pure water___

The Common Ion EffectCalculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10


116


Solution

[Ba2+] [SO42-]

initialchange equilib


+ y0010 0

+ y

0010 + y y

Calculate the solubility of BaSO4 in (a) pure water and (b) in 0010 M Ba(NO3)2Ksp for BaSO4 = 11 x 10-10


117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)

Because y lt 11 x 10-5 M (= x the solubility in pure water) this means 0010 + y is about equal to 0010 Therefore

Ksp = 11 x 10-10 = (0010)(y)

y = 11 x 10-8 M = solubility in presence of added Ba2+ ion



118


SUMMARYSolubility in pure water = x = 11 x 10-5 MSolubility in presence of added Ba2+

= 11 x 10-8 MLe Chatelierrsquos Principle is followed



119


Common Ion Effect PracticeDo you expect the solubility of AgI to be greater in pure water or in 0020 M AgNO3 Explain your answer

Ksp(AgI) = 85 x 10-17

AgI(s) Ag+(aq) + I-(aq)

120


Factors Affecting Solubility - pH

bull pHndash If a substance has a

basic anion (like OH-) it is more soluble in a more acidic solution

ndash Substances with acidic cations (Al3+) are more soluble in more basic solutions

121


Solubility in more Acidic Solution example solid Mg(OH)2 in water

Equilibrium Mg(OH)2 Mg2+ + 2OH-

For Ksp = 18 x 10-11 calculate pH

18 x 10-11 = x(2x)2

x = 165 x 10-4 M [solubility of Mg(OH)2]

= [Mg2+] in solution

[OH-] = 2 x solubility of Mg(OH)2

pOH = -log (2 middot 165 x 10-4) = 348pH = 1052

122


Solubility in more Acidic Solution example solid Mg(OH)2 in pH=9

bufferbull Now put Mg(OH)2 in a pH = 9 buffer solution bull Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)bull [Mg2+][OH-]2 = 18 x 10-11 (Ksp a function of Temp

only)bull [Mg2+] = 18x10-11(1x10-5)2 = 018 Mbull Considerably more soluble (~1000x) in more

acidic solutionbull Solubility in water was 165 x 10-4 Mbull Essentially removed OH- by reacting with H+

123


Acidic solution example Ag3PO4

bull Ag3(PO4) is more soluble in acid solution than neutral solution Why

ndash H+ of acid solution reacts with PO43- anion to

produce weak acid HPO42-

Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = 18x10-18

H+(aq) + PO43- (aq) HPO4

2-(aq) K = 1Ka = 28x1012

Ag3PO4(s) + H+ 3Ag+(aq) + HPO42(aq) K = 50 x 10-6

ndash Addition of H+ shifts equilibrium of 1st rxn to right

ndash PO43- is actually a moderately strong base

PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124


Acidic solution example AgClbull What about AgCl in an acidic solution such

as HNO3 or H2SO4AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 18x10-10

H+(aq) + Cl- (aq)larr HCl(aq) K = very small

bull K of reaction remains very small largely determined by the Ksp of AgCl

bull Note Special case of HCl bull HCl will affect solubility How Whybull It would add chloride actually make AgCl less

soluble by Le Chatelier

125


Solubility and Aciditybull Other conjugate base anions that will increase in

solubility in an acidic solution

OH- F- S2- CO32- C2O4

2- and CrO42-

Why Because all form weak acids

Examples

PbS(s) + 2H+(aq) H2S(g) + Pb2+(aq)

Ca(OH)2(s) + 2H+(aq) 2H2O(l) + Ca2+(aq)

BaCO3(s) + 2H+(aq) CO2(g) + H2O(l) + Ba2+(aq)

126


Precipitation and Qualitative AnalysisSection 162


Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq)

Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

If [Hg22+] = 0010 M what [Cl-] is reqrsquod to just

begin the precipitation of Hg2Cl2

That is what is the maximum [Cl-] that can be

in solution with 0010 M Hg22+ without

forming Hg2Cl2

127


Precipitation and Qualitative Analysis



Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that

Ksp = product of maximum ion concs

Precip begins when product of

ion concentrations (Q)

EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution

[Cl-] that can exist when [Hg22+] = 0010 M

[Cl- ] =

Ksp

0010 = 11 x 10-8 M

If this conc of Cl- is just exceeded Hg2Cl2 begins to precipitate

129




Ksp = 11 x 10-18

Now raise [Cl-] to 10 M What is the value of [Hg2

2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M

The concentration of Hg22+ has been reduced

by 1016

130


Precipitation and Qualitative Analysisndash Practice

bull If a solution is 0020 M in Cl- ions at what concentration of Pb2+ does the PbCl2 (Ksp= 17 x 10-5) precipitate

131


Separating Metal Ions Cu2+ Ag+ Pb2+


Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132


Separating Salts by Differences in Ksp


bull Add CrO42- to solid PbCl2 The less

soluble salt PbCrO4 precipitates

bull PbCl2(s) + CrO42- PbCrO4 + 2 Cl-

bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133



PbCl2(s) + CrO42- PbCrO4 + 2 Cl-

Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

PbCl2(s) Pb2+ + 2 Cl- K1 = Ksp

Pb2+ + CrO42- PbCrO4 K2 = 1Ksp

Knet = (K1)(K2) = 94 x 108

Net reaction is product-favored

134



The color of the salt silver chromate Ag2CrO4 is red A student adds a solution of Pb(NO3)2 to a test tube containing solid red Ag2CrO4 After stirring the contents of the test tube the student finds the contents changes from red to yellowWhat is the yellow salt Which one has a higher Ksp

135


Separating Salts by Differences in KspSolution

136


Separations by Difference in Ksp

137


Qualitative AnalysisFigure 162Group 1 ndash Insoluble ChloridesGroup 2 ndash Sulfides Insoluble in acidGroup 3 ndash Sulfides insoluble in baseGroup 4 ndash Insoluble CarbonatesGroup 5 ndash Alkali Metal and ammonium ions

138


The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to COMPLEX

IONS

Equilibria Involving Complex Ions

Chapter 163

139


Reaction of NH3 with Cu2+

(aq)

>

140


Solubility and Complex Ionsbull Consider the formation of Ag(NH3)2

+

Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)

bull The Ag(NH3)2+ is called a complex ion

bull A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge It happens a lot with transition metals because they have available d-orbitals Bases like ammonia are attracted to that

(See Chapter 21 for details)

141


Solubility and Complex Ionsbull

bull Note that the equilibrium expression is lsquooppositersquo to what we usually write the individual species are on the left and the new substance on the right This leads to Kfrsquos that are very LARGE

142


Solubility and Complex Ions

143


Formation of complex ions explains why you can dissolve a ppt by forming a complex ion

Dissolving Precipitates by forming Complex Ions

AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-

144



bull Complex Ionsndash The formation

of these complex ions increase the solubility of these salts

145


Determine the Keq of a reaction if the solution contains AgBr and NH3bull In water solid AgBr exists in equilibrium with its

ions according to

AgBr Ag+ + Br- Ksp = 5 x 10-13

bull Also Ag+ make complex ions with ammonia

expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107

Solubility and Complex Ions Example

146


bull Therefore the equilibrium expressed by both reactions is


Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107

AgBr + 2NH3 Ag(NH3)2+ + Br- Keq = 85 x 10-6

Keq = Ksp x Kf

bull So this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water

bull In other words NH3 makes AgBr more soluble


147


AP Exam PracticeKsp Problemsbull 2011B AP Exam 1bull 2010 AP Exam 1bull 2006 AP Exam 1bull 2004 AP Exam 1bull 2001 AP Exam 1

Complex AB Problemsbull 2011 AP Exam 1bull 2006B AP Exam 1bull 2003 AP Exam 1bull 2002B AP Exam 1bull 2002 AP Exam 1

More About Chemical Equilibria Acid-Base amp Precipitation React


Stomach Acidity amp Acid-Base Reactions

Acid-Base Reactions

The Common Ion Effect Section 151

Slide 6

Slide 7

Slide 8


pH of NH3NH4+ Mixture (2)

Common Ion Effect


Common Ion Effect Practice (2)

Buffered Solutions Section 152

Buffer Solutions

Buffer Solutions (2)







Henderson-Hasselbalch Equation (2)

Buffer Calculation Example (AcidConj Base)



Buffer Calculation using the HendersonndashHasselbalch Equation (2)

Buffer Practice 1

Buffer Practice 2

Adding an Acid to a Buffer (How a Buffer Maintains pH)



Adding an Acid to a Buffer (2)






Adding a Base to a Buffer (2)




Preparing a Buffer

Preparing a Buffer (2)



Commercial Buffers


Buffer Example

Buffer Example (2)

Preparing a buffer Practice

Buffering Capacity Chapter 153

Buffer Capacity

Large Buffer System


Small Buffer System


Calculate pH via Henderson-Hasselbalch Equations (2)

Mixing Acids and Bases (Stoichiometry amp Equilibrium) Chapter 15

Mixing Acids and Bases (Stoichiometry amp Equilibrium)

Mixing Acids and Bases (Stoichiometry amp Equilibrium) (2)






Slide 67


Titrations and pH Curves (2)


Slide 71




Slide 75






Slide 81

Slide 82






Weak diprotic acid (H2C2O4) titrated with a strong base (NaOH)

Slide 89

Weak base (NH3) titrated with a strong acid (HCl)

Acid-Base Indicators Chapter 155

Acid-Base Indicators



Solubility and Complex Ion Equilibria Chapter 16




Analysis of Silver Group (2)




Solubility Products

Slide 104

Lead(II) Chloride


Solubility of Lead(II) Iodide (2)


SolubilityKsp Practice 1



Common Ion Effect

Barium Sulfate Ksp = 11 x 10-10

The Common Ion Effect (2)







Solubility in more Acidic Solution example solid Mg(OH)2 in wa

Solubility in more Acidic Solution example solid Mg(OH)2 in p


Acidic solution example AgCl

Solubility and Acidity

Precipitation and Qualitative Analysis Section 162


Precipitation and Qualitative Analysis (2)





Separating Salts by Differences in Ksp (2)




Qualitative Analysis

Slide 138

Reaction of NH3 with Cu2+(aq)


Solubility and Complex Ions (2)


Slide 143



Solubility and Complex Ions Example (2)

AP Exam Practice

2


Objectives ndash Chapter 15bull Define the Common Ion Effect (151)bull Define buffer and show how a buffer

controls pH of a solution (152 153)bull Identify and Evaluate titration curves

(154)bull Use of Indicators (155)

3



4


Acid-Base Reactions



HOAc + NH3

What is relative pH

before during amp

after reaction

Need to study

a) Common ion

effect and buffers

b) Titrations

5





equilibrium



NH4+ is an acid


Section 151

6



[NH3] [NH4+] [OH-]




pH of Aqueous NH3

7




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x2

025 - x

pH of Aqueous NH3


[OH-] = x = [Kb(025)]12 = 00021 M


and so pH = 1400 - 267

= 1133 for 025 M NH3

8





[NH3] [NH4+] [OH-]



9





[NH3] [NH4+] [OH-]



10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x



[OH-] = x = (025 010)(Kb) = 45 x 10-5 M



11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

3



4


Acid-Base Reactions



HOAc + NH3

What is relative pH

before during amp

after reaction

Need to study

a) Common ion

effect and buffers

b) Titrations

5





equilibrium



NH4+ is an acid


Section 151

6



[NH3] [NH4+] [OH-]




pH of Aqueous NH3

7




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x2

025 - x

pH of Aqueous NH3


[OH-] = x = [Kb(025)]12 = 00021 M


and so pH = 1400 - 267

= 1133 for 025 M NH3

8





[NH3] [NH4+] [OH-]



9





[NH3] [NH4+] [OH-]



10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x



[OH-] = x = (025 010)(Kb) = 45 x 10-5 M



11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

4


Acid-Base Reactions



HOAc + NH3

What is relative pH

before during amp

after reaction

Need to study

a) Common ion

effect and buffers

b) Titrations

5





equilibrium



NH4+ is an acid


Section 151

6



[NH3] [NH4+] [OH-]




pH of Aqueous NH3

7




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x2

025 - x

pH of Aqueous NH3


[OH-] = x = [Kb(025)]12 = 00021 M


and so pH = 1400 - 267

= 1133 for 025 M NH3

8





[NH3] [NH4+] [OH-]



9





[NH3] [NH4+] [OH-]



10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x



[OH-] = x = (025 010)(Kb) = 45 x 10-5 M



11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

5





equilibrium



NH4+ is an acid


Section 151

6



[NH3] [NH4+] [OH-]




pH of Aqueous NH3

7




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x2

025 - x

pH of Aqueous NH3


[OH-] = x = [Kb(025)]12 = 00021 M


and so pH = 1400 - 267

= 1133 for 025 M NH3

8





[NH3] [NH4+] [OH-]



9





[NH3] [NH4+] [OH-]



10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x



[OH-] = x = (025 010)(Kb) = 45 x 10-5 M



11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

6



[NH3] [NH4+] [OH-]




pH of Aqueous NH3

7




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x2

025 - x

pH of Aqueous NH3


[OH-] = x = [Kb(025)]12 = 00021 M


and so pH = 1400 - 267

= 1133 for 025 M NH3

8





[NH3] [NH4+] [OH-]



9





[NH3] [NH4+] [OH-]



10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x



[OH-] = x = (025 010)(Kb) = 45 x 10-5 M



11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

7




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x2

025 - x

pH of Aqueous NH3


[OH-] = x = [Kb(025)]12 = 00021 M


and so pH = 1400 - 267

= 1133 for 025 M NH3

8





[NH3] [NH4+] [OH-]



9





[NH3] [NH4+] [OH-]



10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x



[OH-] = x = (025 010)(Kb) = 45 x 10-5 M



11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

8





[NH3] [NH4+] [OH-]



9





[NH3] [NH4+] [OH-]



10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x



[OH-] = x = (025 010)(Kb) = 45 x 10-5 M



11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

9





[NH3] [NH4+] [OH-]



10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x



[OH-] = x = (025 010)(Kb) = 45 x 10-5 M



11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

10




Kb= 18 x 10-5 = [NH4

+ ][OH- ]

[NH3 ] =

x(010 + x)

025 - x



[OH-] = x = (025 010)(Kb) = 45 x 10-5 M



11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

11










12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

12





13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

13





With salt

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

14






2-

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

15





H2PO4- + HPO4

2-

NH4+

+ NH3

Buffer Solutions

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

16



We know that


has Kb = 56 x 10-10




Buffer Solutions

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

17





has Ka = 18 x 10-5




Buffer Solutions

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

18




Ka = 18 x 10-5

Buffer Solutions



1 pH lt 245

2 pH gt 245

3 pH = 245



1 pH lt 245

2 pH gt 245

3 pH = 245

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

19






Ka = 18 x 10-5

Buffer Solutions

0700 0600 0-x +x +x

0700 - x 0600 + x x

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

20




Ka= 18 x 10-5 =

[H3O+ ](0600)

0700



Ka = 18 x 10-5

Buffer Solutions

[H3O+] = 21 x 10-5 and pH = 468

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

21





Buffer Solutions



[conjugate base]

[H3O+ ] =

[Acid]

[Conj base] x Ka

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

22





or


23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

23




Solve for [OH-] =



119901119874119867=119901119870 b+119949119952119944 iquestiquest

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

24





25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

25



HC3H5O3 + H2O C3H5O3- + H3O+

I 012 M 010 - E 012-x 010 + x x

x(010 + x) = 14x10-4


x = 012(140 x 10-4) = 168 x 10-4 = [H3O+]

010 So the pH = 377

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

26



pH = ndashlog (14 10-4) + log(010)(012)

pH = 385 + (ndash008)


HC3H5O3 + H2O C3H5O3- + H3O+


27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

27







28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

28


Buffer Practice 1


(Ka = 18 x 10-5) Use H-H Eqn

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

29





HC7H5O2 + H2O H3O+ + C7H5O2-

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

30






water


(100 M)(10 mL) = M2(1001 mL)

M2 = 999 x 10-4 M = [H3O+]

pH = 300

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

31




32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

32








[OAc-] = 0600 M (pH before = 468)

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

33







[OAc-] = 0600 M (pH = 468)

000100 mol0600 mol 0700 mol

-000100 -000100 +000100

0 0599 mol 0701 mol

0 0599 mol1001 L

0701 mol 100l L

0598 molL0700 molL

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

34








[OAc-] = 0600 M (pH = 468)

0700 molL0 0598 molL-x +x +x

0598 + xx0700-x

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

35









[OAc-] = 0600 M (pH = 468)

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

36






[OAc-] = 0600 M (pH = 468)


37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

37





water


(100 M)(10 mL) = M2(1001 mL)

M2 = 100 x 10-3 M = [OH-]

pOH = 300 pH = 1100

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

38








[OAc-] = 0600 M (pH before = 468)

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

39





= 0600 M (pH = 468)

000100 mol0700 mol 0600 mol

-000100 -000100 +000100

0 0699 mol 0601 mol0 0699 mol

1001 L0601 mol 1001 L

0698 molL0600 molL

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

40








= 0600 M (pH = 468)

0698 molL0 0600 molL-x +x +x

0600 + xx0698-x

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

41









= 0600 M (pH = 468 pOH = 932)

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

42






= 0600 M (pH = 468)


43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

43






[H3O+ ] =

[Acid]

[Conj base] x Ka

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

44



[H3O+] = 50 x 10-5 M

POSSIBLE


HSO4- SO4

2- 12 x 10-2

HOAc OAc- 18 x 10-5

HCN CN- 40 x 10-10


Preparing a Buffer

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

45



[H3O+] = 50 x 10-5 M

[H3O+ ] = 50 x 10-5 = [HOAc]

[OAc- ] (18 x 10-5 )



278

1

Preparing a Buffer

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

46




each


Preparing a Buffer

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

47


Commercial Buffers



48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

48




weak acid

160 g Na2CO3 (160106 gmol)

conjugate base

HCO3- + H2O

H3O+ + CO32-

What is the pH

HCO3- pKa = 103

pH = 103 + log (15110)=105

Preparing a Buffer

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

49





2-(aq) + H3O+(aq)


[HPO42-][H2PO4

-] ratio be



50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

50




2-(aq) + H3O+(aq)


-]

740 = 721 + log[HPO42-][H2PO4

-]

log[HPO42-][H2PO4

-] = 019

[HPO42-][H2PO4

-] = 10019 = 155

Part B

60g NaH2PO4 = 0050 mol in 0500 L = 010 M

120gmol

[HPO42-] = 155[H2PO4

-] = 0155 M HPO42-


= 110 g Na2HPO4

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

51



52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

52









53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

53


Buffer Capacity





54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

54


Large Buffer System


H3O+ OAc- HOAc




55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

55




0501 X 499 [base]pH = pKa +

[acid]

pH = -log (18 x 10-5) + log[ 499501]

pH = 4738


56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

56


Small Buffer System


H+ OAc- HOAc




57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

57





Initially 006 0 004

Change -x +x +x

At Equilibrium

006-x x 004 + x

HOAc H3O+ + OAc-

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

58




006-x X 004-x


[base]pH = pKa +

[acid]



59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

59





H+ + OH- H2O(l)


60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

60




H+ + OH- H2O(l)


61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

61




H+ + OH- H2O(l)


pOH = 177 so pH = 1223

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

62






63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

63






64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

64





E 0050-x x x

Krxn =Kb = KwKa = 1x10-1418x10-5 = 556x10-10

= 556x10-10


65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

65






66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

66






67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

67



pHpH


68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

68




slowly

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

69




>

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

70




point

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

71











72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

72






73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

73







74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

74





equivalence point

Bz- + H2O HBz + OH-

Kb = 16 x 10-10

++

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

75







76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

76






77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

77





78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

78







125 mL

[Bz-] = 00025 mol 0125 L = 0020 M

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

79





Bz- + H2O HBz + OH- Kb = 16 x 10-10



80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

80





Bz- + H2O HBz + OH- Kb = 16 x 10-10


x = [OH-] = 18 x 10-6

pOH = 575 and pH = 825

Kb = 16 x 10-10 =

x2

0020 - x

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

81








82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

82






pH = 825


83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

83





[H3O+ ] = [HBz]

[Bz- ] x Ka




This is a BUFFER


This is a BUFFER

HBz + H2O H 3O+ + Bz- Ka = 63 x 10-5

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

84






OH- + HBz H2O + Bz-

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

85


OH- + HBz H2O + Bz-







86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

86




87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

87


See Figure 184


88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

88



(NaOH)

See Figure 186

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

89



pH


90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

90



acid (HCl)

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

91








92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

92



93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

93





94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

94





95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

95



Chapter 16


Chapter 16

Lead(II) iodide

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

96







97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

97









98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

98





Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

99


Analysis of Silver

Group

Analysis of Silver



SLIGHT extent




Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

100


Analysis of Silver

Group

Analysis of Silver

Group




What is [Cl-]

[Cl-] = [Ag+] = 167 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

101


Analysis of Silver

Group

Analysis of Silver



[Ag+] = [Cl-] = 167 x 10-5 M


Kc = [Ag+] [Cl-]

= (167 x 10-5)(167 x 10-5)

= 279 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

102


Analysis of Silver

Group

Analysis of Silver

Group


Kc = [Ag+] [Cl-] = 279 x 10-10



constant



Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

103





Ksp =[M+]m[A-]a


104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

104


Some Values of Ksp

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

105



Ksp = 19 x 10-5 = [Pb2+][Clndash]2

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

106


Solution

1 Solubility = [Pb2+] = 130 x 10-3 M

[I-] =

[I-] = 2 x [Pb2+] = 260 x 10-3 M





Calculate Ksp


107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

107


Solution

2 Ksp = [Pb2+] [I-]2

= [Pb2+] 2 bull [Pb2+]2

Ksp = 4 [Pb2+]3




Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9



Calculate Ksp


108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

108


Caveat








Ksp = 4 (130 x 10-3)3 = 879 x 10-9Ksp = 4 (130 x 10-3)3 = 879 x 10-9

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

109





110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

110




111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

111




112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

112



Ksp = 19 x 10-5

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

113


Barium

Sulfate

Ksp = 11 x 10-10



114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

114





Solution


Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)12 = 11 x 10-5 M



115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

115







116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

116


Solution

[Ba2+] [SO42-]



+ y0010 0

+ y

0010 + y y



117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

117


Solution

Ksp = [Ba2+] [SO42-] = (0010 + y) (y)


Ksp = 11 x 10-10 = (0010)(y)




118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

118






119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

119



Ksp(AgI) = 85 x 10-17


120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

120






121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

121





18 x 10-11 = x(2x)2





122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

122






123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

123








2-(aq) K = 1Ka = 28x1012




PO43- + H2O HPO4

2- + OH- Kb3 = 28x10-2

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

124








125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

125




OH- F- S2- CO32- C2O4

2- and CrO42-


Examples




126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

126





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2





forming Hg2Cl2

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

127





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Recognize that




EXCEEDS the Ksp

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

128





Ksp = 11 x 10-18 = [Hg22+] [Cl-]2

Solution


[Cl- ] =

Ksp

0010 = 11 x 10-8 M


129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

129




Ksp = 11 x 10-18


2+] at this point

Solution

[Hg22+] = Ksp [Cl-]2

= Ksp (10)2 = 11 x 10-18 M


by 1016

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

130




131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

131




Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

Ksp Values

AgCl 18 x 10-10

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

132







bull Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

133




Salt Ksp

PbCl2 17 x 10-5

PbCrO4 18 x 10-14



Knet = (K1)(K2) = 94 x 108


134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

134




135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

135



136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

136



137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

137



138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

138



IONS


Chapter 163

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

139



(aq)

>

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

140



+





141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

141




142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

142



143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

143





144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

144





145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

145



ions according to



expressed as

Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

146




Ag+ + 2NH3 Ag(NH3)2+ Kf = 17 x 107


Keq = Ksp x Kf




147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

147







Acid-Base Reactions


Slide 6

Slide 7

Slide 8



Common Ion Effect




Buffer Solutions













Buffer Practice 1

Buffer Practice 2














Preparing a Buffer




Commercial Buffers


Buffer Example

Buffer Example (2)



Buffer Capacity

Large Buffer System


Small Buffer System











Slide 67




Slide 71




Slide 75






Slide 81

Slide 82







Slide 89














Solubility Products

Slide 104

Lead(II) Chloride







Common Ion Effect


























Slide 138





Slide 143




AP Exam Practice

1 © 2009 Brooks/Cole - Cengage More About Chemical Equilibria Acid-Base & Precipitation Reactions...

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Transcript of 1 © 2009 Brooks/Cole - Cengage More About Chemical Equilibria Acid-Base & Precipitation Reactions...