16-1 CHEM 102, Fall 2014, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected] Office:...

16-1CHEM 102, Fall 2014, LA TECH

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

Office: CTH 311

Phone 257-4941

Office Hours: M,W 8:00-9:00 & 11:00-12:00 am;

Tu,Th,F 9:30 - 11:30 am. or by appointment.

Test Dates: 9:30-10:45 am., CTH 328

Chemistry 102(01) Spring 2014

March 31, 2014 (Test 1): Chapter 13

April 23, 2014 (Test 2): Chapter 14 &15

May 19, 2014 (Test 3) Chapter 16 &17

May 21, 2014 (Make-up test) comprehensive:

Chapters 13-17


Chapter 16.Aditional Aqueous Equilibria17.1 Buffer Solutions

17.2 Acid-Base Titrations

17.3 Acid Rain

17.4 Solubility Equilibria and the Solubility Product Constant, Ksp

17.5 Factors Affecting Solubility / Precipitation: Will It Occur?


Reaction of a basic anion or acidic cation with water is an ordinary Brønsted-Lowry acid-

base reaction.

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

NH4+

(aq) + H2O(l) NH3 (aq) + H3O+

(aq)

This type of reaction is given a special name.

Hydrolysis

The reaction of an anion with water to produce the conjugate acid and OH-.

The reaction of a cation with water to produce the conjugate base and H3O+

.

Hydrolysis


Acid-Base Properties of Typical Ions


What salt solutions would be acidic, basic and neutral?

1) strong acid + strong base = neutral

2) weak acid + strong base = basic

3) strong acid + weak base = acidic

4) weak acid + weak base = neutral,

basic or an acidic solution depending on the relative strengths of the acid and the base.


What pH? Neutral, basic or acidic?

•a)NaCl • neutral•b) NaC2H3O2

• basic•c) NaHSO4 • acidic•d) NH4Cl• acidic


1) If the following substance is dissolved in pure water, will the solution be acidic, neutral, or basic?

a) Solid sodium carbonate-(Na2CO3):

b) Sodium chloride- (NaCl):

c) Sodium acetate- (NaC2H3O2):

d) Ammonium sulfate-((NH4)2SO4):


How do you calculate pH of a salt solution?Find out the pH, acidic or basic?

If acidic it should be a salt of weak base

If basic it should be a salt of weak acid

if acidic calculate Ka from Ka= Kw/Kb

if basic calculate Kb from Kb= Kw/Ka

Do a calculation similar to pH of a weak acid or base


What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5)

Find out the pH, acidic

if acidic calculate Ka from Ka= Kw/Kb

Ka= Kw/Kb = 1 x 10-14 /1.8 x 10-5)

Ka= 5.56. X 10-10

Do a calculation similar to pH of a weak acid


Continued

NH4+ + H2O H 3

+O + NH3

[NH4+] [H3

+O ] [NH3 ]Ini. Con. 0.5 M 0.0 M 0.00 MChange -x x xEq. Con. 0.5 - x x x [H 3+O ] [NH3 ]

Ka(NH4+) = -------------------- =

[NH 4+] x2

---------------- ; appro.:0.5 - x . 0.5 (0.5 - x)


x2 Ka(NH4

+) = ----------- = 5.56 x 10 -10

0. 5 x2

= 5.56 x 10 -10 x 0.5 = 2.78 x 10 -10

x= 2.78 x 10 -10 = 1.66 x 10-5

[H+ ] = x = 1.66 x 10-5 MpH = -log [H+ ] = - log 1.66 x 10-5

pH = 4.77

pH of 0.5 M NH4Cl solution is 4.77 (acidic)

Continued


2) What is the pH of a 0.05 M aqueous NH4Cl solution? (Kb (NH3) = 1.8 x 10-5)

a) equilibrium reaction for the hydrolysis of salt:

b) Ka for the conjugate acid NH4+:

c) ICE set-up:

I:___________________________________________

C:__________________________________________

E___________________________________________

d) Calculation of x, [H3O]+:

e) pH of the solution:


3) What is the pH of a 0.05 M aqueous NaC2H3O2 solution? (Ka (HC2H3O2) = 1.8 x 10-5)

a) equilibrium reaction for the hydrolysis of salt:

b) Kb for the conjugate base C2H3O2-:

c) ICE set-up:

I:___________________________________________

C:__________________________________________

E___________________________________________

d) Calculation of x, [OH]-:

e) pOH and pH of the solution:


Acid-Base Chemistryof Some Antacids


4) A 50.00-mL sample of 0.100 M KOH is being titrated with 0.100 M HNO3. Calculate the pH of the solution after 52.00 mL of HNO3 is added.

a) acid base reaction:

b) moles of KOH: c) moles of HNO3:

d) [H3O]+:



Reaction of a basic anion with water is an ordinary Brønsted-Lowry acid-base reaction.

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

This type of reaction is given a special name.

Hydrolysis

The reaction of an anion with water to produce the conjugate acid and OH-.

The reaction of a cation with water to produce the conjugate base and H3O+.

Hydrolysis


Common Ion Effect

Weak acid and salt solutions

E.g. HC2H3O2 and NaC2H3O2

Weak base and salt solutions

E.g. NH3 and NH4Cl.

H2O + C2H3O2- OH- + HC2H3O2

(common ion)

H2O + NH4 + H3+O + NH3

(common ion)


Solutions that resist pH change when small amounts of acid or base are added.

Two types

Mixture of weak acid and its salt

Mixture of weak base and its salt

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Add OH- Add H3O+

shift to right shift to left

Based on the common ion effect.

Buffers


The pH of a buffer does not depend on the absolute amount of the conjugate acid-base

pair. It is based on the ratio of the two.

Henderson-Hasselbalch equation.

Easily derived from the Ka or Kb expression.

Starting with an acid

pH = pKa + log

Starting with a base

pH = 14 - ( pKb + log )[HA]

[A-]

[A-]

[HA]

Buffers


Henderson-Hasselbalch Equation

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[H3O+] [A-] Ka = ---------------- [HA]

[H3O+] = Ka ([HA]/[A-])

pH = pKa + log([A-]/[HA])

when the [A-] = [HA]pH = pKa


Calcualtion of pH of BuffersHenderson Hesselbach Equation

[ACID]

pH = pKa - log ---------

[BASE]

[BASE]

pH = pKa + log ---------

[ACID]


Control of blood pH

Oxygen is transported primarily by hemoglobin in the red blood cells.

CO2 is transported both in plasma and the red blood cells.

CO2 (aq) + H2O H2CO3 (aq)

H+

(aq) + HCO3-(aq)

The bicarbonate

buffer is essential

for controlling

blood pH

The bicarbonate

buffer is essential

for controlling

blood pH

Buffers and blood


Buffer Capacity

Refers to the ability of the buffer to retard changes in pH when small amounts of acid or base are added

The ratio of [A-]/[HA] determines the pH of the buffer whereas the magnitude of [A-] and [HA] determine the buffer capacity


Adding an Acid or Baseto a Buffer


Buffer Systems


5) For the (buffer effect) of HC2H3O2/NaC2H3O2

a) Acid dissociation reaction:

b) Salt hydrolysis reaction:

c) Common ions in both equilibria:

d) Which way salt hydrolysis equilibrium move adding

H3O+:

e) Which way salt hydrolysis equilibrium move adding

OH-:


6) Describe the (buffer effect) of NH3/NH4Cl

a) Buffer type: (weak acid or base)/soluble salt):

b) Base dissociation reaction:

c) Salt hydrolysis reaction:

d) common ions in both equilibria:

e) Which way salt hydrolysis equilibrium move adding

H3O+:

f) Which way salt hydrolysis equilibrium move adding

OH-:


7) What is the pH of a solution that is 0.2 M in acetic acid (Ka = 1.8 x 10-5) and 0.2 M in sodium acetate?

a) Is it a acid, base, salt or buffer solution?

b) Henderson-Hesselbalch equation:

c) pKa: d)



Titrations ofAcids and Bases

Titration

Analyte

Titrant

analyte + titrant => products


Acid-base indicators are highly colored weak acids or bases.

HIn In- + H+

color 1 color 2

They may have more than one color transition.

Example. Thymol blue

Red - Yellow - Blue

One of the forms may be colorless - phenolphthalein (colorless to pink)

Indicators


Acid-Base Indicator

HIn + H2O H3O+ + In-

acid base

color color

[H3O+][In-]Ka =

[HIn]They may have more than one color transition.

Example. Thymol blue

Red - Yellow - Blue

Weak acid that changes color with changes in pH


What is an Indicator?Indicator is an weak acid with different Ka, colors to

the acid and its conjugate base. E.g.

phenolphthalein Hin H+ + In-

colorless pink

Acidic colorless

Basic pink


Selection of an indicator for a titration

a) strong acid/strong base

b) weak acid/strong base

c) strong acid/weak base

d) weak acid/weak base

Calculate the pH of the solution at he equivalence point or end point


pH and Color of Indicators


Red Cabbage as IndicatorC O H

O

N N N

CH3

CH3

(aq)

C O-

O

N N N

CH3

CH3

(aq) + H3O+ (aq)

yellow

red

+ H2O (l)


Acid-base indicators are weak acids that undergo a color change at a

known pH.

phenolphthalein

pH

Indicator examples


8) If 50 ml of a 0.01 M HCl solution is titrated with a 0.01 M NaOH solution, what will be the concentration of salt (NaCl) the pH at the endpoint?

a) NaCl solution acidic, basic or neutral?

b) Concentration of [NaCl]:

c) pH of the solution?

d) Suitable indicator for the titration:


Titration Apparatus

Burette delivering base to a flask containing an acid.

The pink color in the flask is due to the

phenolphthalein indicator.


Endpoint vs. Equivalence Point

Endpointpoint where there is a physical change, such

as color change, with the indicator

Equivalence Point# moles titrant = # moles analyte

#molestitrant=(V M)titrant

#molesanalyte=(V M)analyte


9) If 50 mL of a 0.01 M HCl solution is titrated with a 0.01 M NH3 (Kb = 1.8 x 10-5) solution, what will be

a) The initial pH (0.01 M NH3):

b) Concentration of NH4Cl at the endpoint:

c) pH at the endpoint:

d) Suitable indicator for the titration:


10) If 50 ml of a 0.01 M HC2H3O2 solution is titrated with a 0.01 M NaOH solution, what will be the

a) Molarity of NaC2H3O2 at the endpoint:

b) The pH at the endpoint:

c) What indicator would be most suitable for this titration:


Polyprotic Acids


Organic or Carboxylic Acids

H C

H

H

C

H

H

C

H

H

C

O

O H

nonacidic hydrogens

butanoic acid

acidic hydrogen

CH 3 C

O

acetic acid

OH

electron-attractingoxygen atom

acidic hydrogen

CH 3 C

O

OH

-

CH 3 C

O

O-

acetate ion


FCH2CO2H (strongest acid) > ClCH2CO2H > BrCH2CO2H (weakest acid).

Acid Ka pKa

HCOOH (formic acid) 1.78 X 10-43 0.75

CH3COOH (acetic acid) 1.74 X 10-54 0.76

CH3CH2COOH (propanoic acid)1.38 x 10-5 4.86

Organic or Carboxylic Acids


Acid-Base in the Kitchenvinegar - acetic acid

lemon juice (citrus juice) - citric acid

baking soda - NaHCO3

milk - lactic acid

baking powder - H2PO4- & HCO3

-


Household Cleaners

CH 3CH2CH 2CH2CH 2CH2CH 2CH 2CH 2CH 2CH2CH 2CH2CH 2SO3

-Na+

Oil-soluble part(hydrophobic)

Water-soluble part(hydrophilic)

A Typical Synthetic Detergent Molecule

CH 3(CH 2)4COO(CH 2)2O( CH2CH 2O) 2CH 2CH 2OH

esterlink

(hydro-philic)

etherlink

etherlink

(hydrophilic)

hydrocarbonchain

(hydro-phobic)

alcohol group(hydrophilic)

A nonionic detergent


Dishwashing Detergent


Acid-Base Indicator Behavioracid color shows when

[In-] 1

£ [HIn] 10

[H3O+

][In-] 1

= [H3O+

] = Ka

[HIn] 10

base color shows when

[In-] 1

³ [HIn] 10

[H3O+][In-]

= 10 [H3O+] = Ka

[HIn]


Indicator pH Range

acid color shows when

pH + 1 = pKa

and base color shows when

pH - 1 = pKa

Color change range is

pKa = pH 1 or pH = pKa 1


Titration curves

Acid-base titration curve

A plot of the pH against the amount of acid or base added during a titration.

Plots of this type are useful for visualizing a titration.

It also can be used to show where an indicator undergoes its color change.


Indicator and Titration Curve 0.1000 M HCl vs 0.1000 M NaOH


EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

at 0.00 mL of NaOH added, initial point

[H3O+

] = [HCl] = 0.1000 M

pH = 1.0000



at 15.00 mL of NaOH added

Va Ma > Vb Mb thus

(Va Ma) - (Vb Mb)

[H3O+] =

(Va + Vb)

((35.00mL) (0.1000M)) - ((15.00mL) (0.1000M))

= (35.00 + 15.00)mL

= 4.000 10-2

M pH = 1.3979




Va Ma = Vb Mb , equivalence point

at equivalence point of a

strong acid - strong base titration

pH º 7.0000




Vb Mb > Va Ma , post equvalence point

(Vb Mb) - (Va Ma)

[OH-] =

(Va + Vb)

((50.00mL) (0.1000M)) - ((35.00mL) (0.1000M))

=

(35.00 + 50.00)mL

= 1.765 10-2

M pOH = 1.7533

pH = 14.00 - 1.7533 = 12.25


0

5

10

15

0 10 20 30 40 50

Volume of Base Added

pH equivalence point x

Titration of Strong Acid with Strong Base


Titration of Weak Acid with Strong Base


Effect of Acid Strength on Titration Curve


Titration of Weak Base with Strong Acid


Titration of Diprotic Weak Acid with Strong Base


pH range of Indicators

litmus (5.0-8.0)bromothymole blue (6.0-7.6) methyl red (4.8-6.0)thymol blue (8.0-9.6) phenolphthalein (8.2-10.0) thymolphthalein (9.4-10.6)


Acid Rain

acid rain is defined as rain with a

pH < 5.6

pH = 5.6 for rain in equilibrium with atmospheric carbon dioxide


Sulfuric Acid from Sulfur burning

SO2

S + O2 => SO2

SO3

2 SO2 + O2 => 2 SO3

Sulfuric Acid

SO3 + H2O => H2SO4


Nitric Acid2 NO2(g) + H2O(l) => HNO3(aq) + HNO2(aq)


How Acid Precipitation Forms


Acid Precipitation in U.S.


Solubility Productsolubility-product - the product of the solubilities

solubility-product constant => Ksp

constant that is equal to the solubilities of the ions produced when a substance dissolves


Solubility Product Constant

In General:

AxBy xA+y + yB-x

[A+y

]x [B-x

]y

K = [AxBy]

[AxBy] K = Ksp = [A+y

]x

[B-x

]y

Ksp = [A+y

]x [B

-x]y

For silver sulfate

Ag2SO4 2 Ag+

+ SO4-2

Ksp = [Ag+

]2

[SO4-2

]


Solubility Product Constant Values


Dissolving Slightly Soluble Salts Using Acids

Insoluble salts containing anions of Bronsted-Lowry bases can be dissolved in solutions of low pH


Calcium CarbonateDissolved in Acid

Limestone Dissolving in Ground Water

CaCO3(S) + H2O + CO2 => Ca+2(aq) + 2 HCO3

-(aq)

Stalactite and Stalagmite Formation

Ca+2(aq) + 2 HCO3

-(aq) => CaCO3(S) + H2O + CO2


The Common Ion Effect

common ion

second source which is completely dissociated

In the presence of a second source of the ion, there will be less dissolved than in its absence

common ion effect

a salt will be less soluble if one of its constitutent ions is already present in the solution


EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl?

AgCl Ag+ + Cl-

Ksp = [Ag+][Cl-] = 1.82 10-10M2

let x = molar solubility = [Ag+] = [Cl-]

(x)(x) = Ksp = [Ag+][Cl-] = 1.82 10-10M2

x = 1.35 10-5M


EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl?

AgCl Ag+ + Cl-

Ksp = [Ag+][Cl-] = 1.82 10-10M2

let x = molar solubility = [Ag+]

[Cl-] = 1.0 M

Ksp = [Ag+][Cl-] = (x)(1.0M) = 1.82 10-10M2

x = 1.82 10-10M


Formation of Complexesligand - Lewis base

complexes - product of Lewis acid-base reaction

Ag+(aq) + 2 NH3(aq) [Ag(NH3)2(aq)]+

Ag+(aq) + Cl-

(aq) AgCl(s)

AgCl(s) + 2 NH3(aq) [Ag(NH3)2(aq)]+ + Cl-(aq)


Sodium Thiosulfate Dissolves Silver Bromide


11) For a saturated solution of

a) AgCl in water:

i) Solubility equilibrium reaction:

ii) Ksp expression:


11) For a saturated solution of

b) CaF2 in water:


ii) Ksp expression:

c) Fe2S3 in water


ii) Ksp expression:


12) Which of following has the highest molar solubility (mole/L)?

Salt Ksp

a) CaCO3 5 × 10-9

b) PbCO 3 1.4 × 10-13

c) Li2CO3 2 × 10-3

d) NiCO3 1.2 × 10-7


Formation Constants


Amphoterism


Reactant Quotient, Q

Reactant Quotient, Q

ion product of the initial concentration

same form as solubility product constant

Q < Ksp - no precipitate forms an unsaturate

solution

Q > Ksp - precipitate may form to restore condition of saturated solution

Q = Ksp - no precipitate forms, saturated

solution


Will Precipitation Occur?


13) For Li2CO3, Ksp is 2 × 10-3 M3. What is the concentration of Li+ in a saturated solution of Li2CO3?


14) Chemical analysis gave [Pb2+] = 0.012 M, and [Br-] = 0.024 M in a solution. From a table, you find Ksp for PbBr2 has a value of 4 x 10-5 M3.

a) Solubility equilibrium reaction:

b) Qsp:

c) Ksp:

d) Qsp < Ksp, Qsp = Ksp or Qsp > Ksp?

e) Is the solution saturated, oversaturated or unsaturated?


Kidney StonesKidney stones are normally composed of:

calcium oxalate

calcium phosphate

magnesium ammonium phsphate


Calcium OxalatePrecipitate formed from calcium ions from food rich

in calcium, dairy products, and oxalate ions from fruits and vegetables

Ca+2 + C2O4-2 CaC2O4


PRECIPITATION REACTIONSPRECIPITATION REACTIONS


Analysis of Silver GroupAnalysis of Silver Group

These salts formed are 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




When solution is SATURATED, expt. shows that [Ag+] = 1.67 x 10-5 M.

(SOLUBILITY) of AgCl.

What is [Cl-]?

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2




Saturated solution has [Ag+] = [Cl-] = 1.67 x 10-5 M

Use this to calculate Ksp

Ksp = [Ag+] [Cl-]

= (1.67 x 10-5)(1.67 x 10-5)

= 2.79 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2


Because this is the product of “solubilities”, we call it

Ksp = solubility product constant

See Table in the Text

Ksp = solubility product


Lead(II) ChlorideLead(II) Chloride

PbCl2(s) Pb2+(aq) + 2 Cl-(aq)

Ksp = 1.9 x 10-5


Consider PbI2 dissolving in water

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

Calculate Ksp if solubility = 0.00130 M

Solution1. Solubility = [Pb2+]

= 1.30 x 10-3 M

[I-] = 2 x [Pb2+] = 2.60 x 10-3 M

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

= [Pb2+] {2 • [Pb2+]}2

= 4 [Pb2+]3 = 4 (solubility)3

Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9

Solubility of Lead(II) IodideSolubility of Lead(II) Iodide


Precipitating an Insoluble SaltPrecipitating an Insoluble SaltHg2Cl2(s) Hg2

2+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

If [Hg22+] = 0.010 M, what [Cl-] is req’d to just begin

the precipitation of Hg2Cl2?

(maximum [Cl-] in 0.010 M Hg22+ without forming

Hg2Cl2?)


Precipitating an Insoluble SaltPrecipitating an Insoluble Salt

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

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

Recognize that

Ksp = product of maximum ion concs.

Precip. begins when product of

ion Concs. EXCEEDS the Ksp.




Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

Solution

[Cl-] that can exist when [Hg22+] = 0.010 M,

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

precipitate.

[Cl ] = Ksp

0.010 = 1.1 x 10-8 M




Ksp = 1.1 x 10-18

Now raise [Cl-] to 1.0 M. What is the value of [Hg2

2+] at this point?

Solution

[Hg22+] = Ksp / [Cl-]2

= Ksp / (1.0)2 = 1.1 x 10-18 M

The concentration of Hg22+ has been reduced

by 1016 !


Separating Metal Ions Cu2+, Ag+, Pb2+

Separating Metal Ions Cu2+, Ag+, Pb2+

Ksp ValuesAgCl 1.8 x 10-10

PbCl2 1.7 x 10-5

PbCrO4 1.8 x 10-14

Ksp ValuesAgCl 1.8 x 10-10

PbCl2 1.7 x 10-5

PbCrO4 1.8 x 10-14

Cu2+

Ag+

Pb2+

Cl-

Insoluble

PbCl2 AgCl

Soluble

CuCl2

Heat

Insoluble

AgCl

Soluble

PbCl2

CrO4-2

Insoluble

PbCrO4


Separating Salts by Differences in KspSeparating Salts by Differences in Ksp

A solution contains 0.020 M Ag+ and Pb2+. Add CrO42-

to precipitate Ag2CrO4 (red) and PbCrO4 (yellow). Which precipitates first?

Ksp for Ag2CrO4 = 9.0 x 10-12

Ksp for PbCrO4 = 1.8 x 10-14

Solution

The substance whose Ksp is first exceeded

precipitates first. The ion requiring the smaller amount of CrO4

2- ppts. first.



[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]

= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M

[CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2

= 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M

PbCrO4 precipitates first.

Solution

Calculate [CrO42-

] required by each ion.


How much Pb2+ remains in solution when Ag+ begins to precipitate?

Solution

We know that [CrO42-] = 2.3 x 10-8 M to begin to ppt. Ag2CrO4.

What is the Pb2+ conc. at this point?

[Pb2+] = Ksp / [CrO42-] = 1.8 x 10-14 / 2.3 x 10-8 M

= 7.8 x 10-7 M

Lead ion has dropped from 0.020 M to < 10-6 M



Common Ion EffectAdding an Ion “Common” to an Equilibrium

Common Ion EffectAdding an Ion “Common” to an Equilibrium

PbCl2(s) Pb2+

(aq) + 2Cl-(aq)

NaCl Na+

(aq) + Cl- (aq)


Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10

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

Solution

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

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

x = (Ksp)1/2 = 1.1 x 10-5 M

The Common Ion EffectThe Common Ion Effect



Ksp = 1.1 x 10-10

Solution

b) Now dissolve BaSO4 in water already containing 0.010 M Ba2+.

Which way will the “common ion” shift the equilibrium? ___

Will solubility of BaSO4 be less than or greater than in pure water?___




Solution

[Ba2+] [SO42-]

initial

change

equilib.



Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10


Solution

[Ba2+] [SO42-]

initial 0.010 0

change + y + y

equilib. 0.010 + y y



Ksp = [Ba2+] [SO42-] = (0.010 + y) (y)

Because y << x (1.1 x 10-5 M) 0.010 + y 0.010. Therefore,

Ksp = 1.1 x 10-10 = (0.010)(y)

y = 1.1 x 10-8 M = solubility in presence of added Ba2+ ion.

Le Chatelier’s Principle is followed!


16-1 CHEM 102, Fall 2014, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected] Office:...

Documents

Transcript of 16-1 CHEM 102, Fall 2014, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected] Office:...