Chapter 15dfard.weebly.com/uploads/1/0/5/3/10533150/chem1b_ch15.pdf · Chapter 15 Applications of...

Chapter 15Chapter 15

Applications of Aqueous EquilibriaApplications of Aqueous Equilibria

Addition of base:

Normal human blood pH is 7.4 and has a narrow range of about +/- 0.2

• When an acid or

base is added to

water, the pH

changes drastically.

• A buffer solution

resists a change in

pH when an acid or

base is added.

Buffer SolutionsBuffer Solutions

Buffers:

•Mixture of a weak acid and its salt

•Mixture of a weak base and its salt

pH = 4

pH = 10.5

pH = 6.9

pH = 7.1

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

The presence of a common ion suppresses

the ionization of a weak acid or a weak base.

Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).

CH3COONa (s) Na+ (aq) + CH3COO- (aq)

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

common ion

Understanding Buffer SolutionsUnderstanding Buffer Solutions

Consider mixture of salt NaA and weak acid HA.

HA (aq) H+ (aq) + A- (aq)

NaA (s) Na+ (aq) + A- (aq)

Ka =[H+][A-]

[HA]

[H+] =Ka [HA]

[A-]

-log [H+] = -log Ka - log[HA][A-]

-log [H+] = -log Ka + log[A-][HA]

pH = pKa + log[A-][HA]

pKa = -log Ka

Henderson-Hasselbalch equation

pH = pKa + log[conjugate base]

[acid]

pH of a Buffer:

Using Henderson-Hasselbalch equationUsing Henderson-Hasselbalch equation

pH = pKa + log[HCOO-][HCOOH]

HCOOH : Ka= 1.67 X 10-4 , pKa = 3.78

pH = 3.77 + log[0.52][0.30]

= 4.02

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (Ka= 1.67 X 10-4 ).

A buffer solution is a solution of:

1. A weak acid or a weak base and

2. The salt of the weak acid or weak base

Both must be present!

A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

Add strong acid

H+ (aq) + CH3COO- (aq) CH3COOH (aq)

Add strong base

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

Consider an equal molar mixture of CH3COOH and CH3COONa

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3

(a) HF is a weak acid and F- is its conjugate basebuffer solution

(b) HBr is a strong acidnot a buffer solution

(c) CO32- is a weak base and HCO3

- is it conjugate acidbuffer solution

Blood Plasma pH = 7.4

Buffer Solutions 02Buffer Solutions 02

= 9.20

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?(Ka= 5.62 X 10-10 ).

NH4+ (aq) H+ (aq) + NH3 (aq)

pH = pKa + log[NH3][NH4

+]pKa = 9.25 pH = 9.25 + log

[0.30][0.36]

= 9.17

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

Initial (moles)

End (moles)

0.36 X0.0 800 =0.029 0.050 x 0.0200 = 0.001 0.30 X .0800 = 0.024

0.029-0.001 = 0.028 0.001-0.001 =0.0 0.024 + 0.001 = 0.025

pH = 9.25 + log[0.25][0.28]

[NH4+] =

0.0280.10

final volume = 80.0 mL + 20.0 mL = 100 mL , 0.1 L

[NH3] = 0.0250.10

-0.001 -0.001 +0.001Change

Titrations

Slowly add baseto unknown acid

UNTIL

The indicatorchanges color

(pink)

TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) theequivalence point

Slowly add baseto unknown acid

UNTIL

The indicatorchanges color

(pink)

Acid–Base Titration Curves 01Acid–Base Titration Curves 01

Strong Acid-Strong Base TitrationsStrong Acid-Strong Base Titrations

Equivalence Point: The point at which stoichiometrically equivalent quantities of acid and base have been mixed together.


1. Before Addition of Any NaOH

0.100 M HCl

2. Before the Equivalence Point


2H2O(l)H3O1+(aq) + OH1-(aq)100%

Excess H3O1+

3. At the Equivalence Point


The quantity of H3O1+ is equal to the quantity of added OH1-.

pH = 7

4. Beyond the Equivalence Point


Excess OH1-

Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations


1. Before Addition of Any NaOH

0.100 M CH3CO2H

H3O1+(aq) + CH3CO21-(aq)CH3CO2H(aq) + H2O(l)


2. Before the Equivalence Point

H2O(l) + CH3CO21-(aq)CH3CO2H(aq) + OH1-(aq)

100%

Excess CH3CO2H

Buffer region

췠ѥ


3. At the Equivalence Point

OH1-(aq) + CH3CO2H(aq)CH3CO21-(aq) + H2O(l)

Notice how the pH at the equivalence point is shifted up versus the strong acid titration.

pH > 7

Notice that the strong and weak acid titration curves correspond after

the equivalence point.


4. Beyond the Equivalence Point

Excess OH1-

Titration of a Weak Acid with a Strong BaseTitration of a Weak Acid with a Strong Base

With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

뫐

Strong Acid-Weak Base Titrations

HCl (aq) + NH3 (aq) NH4Cl (aq)

NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)

At equivalence point (pH < 7):

H+ (aq) + NH3 (aq) NH4+ (aq)

Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 MNaOH solution. What is the pH at the equivalence point ?

HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)

start (moles)

end (moles)

0.01 0.01

0.0 0.0 0.01

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00

+x

x x

[NO2-] =

0.010.200 = 0.05 MFinal volume = 200 mL

Kb =[OH-][HNO2]

[NO2-]

=x2

0.05-x= 2.2 x 10-11

0.05 – x ≈ 0.05 x ≈ 1.05 x 10-6 = [OH-]

pOH = 5.98

pH = 14 – pOH = 8.02

Polyprotic Acids:Titration of Diprotic Acid + Strong Base

Polyprotic Acids:Titration of Diprotic Acid + Strong Base

ѫ

Titration of a Triprotic Acid: H3PO4 + Strong BaseTitration of a Triprotic Acid: H3PO4 + Strong Base

터

• Fractional precipitation is a method of removing one ion

type while leaving others in solution.

• Ions are added that will form an insoluble product with one

ion and a soluble one with others.

• When both products are insoluble, their relative Ksp values

can be used for separation.

Fractional Precipitation 01Fractional Precipitation 01

터

Solubility Equilibria

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

Ksp = [Ag+][Cl-] Ksp is the solubility product constant

MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO4

3-]2

Dissolution of an ionic solid in aqueous solution:

Q (IP) = Ksp Saturated solution

Q (IP) < Ksp Unsaturated solution No precipitate

Q (IP) > KspSupersaturated solution Precipitate will form

Ksp = 1.8 x 10-10

Ksp : Solubility Product(IP) : Ion product

터

Measuring Ksp and Calculating Solubility from Ksp

Measuring Ksp and Calculating Solubility from Ksp

터

Molar solubility (mol/L): Number of moles of a solute that

dissolve to produce a litre of saturated solution

Solubility (g/L) is the number of grams of solute that dissolve

to produce a litre of saturated solution

터

Measuring KspMeasuring Ksp

If the concentrations of Ca2+(aq) and F1-(aq) in a saturated solution of calcium fluoride are known, Ksp may be calculated.

Ca2+(aq) + 2F1-(aq)CaF2(s)

Ksp = [Ca2+][F1-]2 = (3.3 x 10-4)(6.7 x 10-4)2 = 1.5 x 10-10

[F1-] = 6.7 x 10-4 M[Ca2+] = 3.3 x 10-4 M

(at 25 °C)

터

What is the solubility of silver chloride in g/L ?

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

Ksp = [Ag+][Cl-]Initial (M)

Change (M)

Equilibrium (M)

0.00

+s

0.00

+s

s s

Ksp = s2

s = Ksp√s = 1.3 x 10-5

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M

Solubility of AgCl = 1.3 x 10-5 mol AgCl

1 L soln143.35 g AgCl

1 mol AgClx = 1.9 x 10-3 g/L

Ksp = 1.8 x 10-10

Calculating Solubility from Ksp

터








터

If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form?

The ions present in solution are Na+, OH-, Ca2+, Cl-.

Only possible precipitate is Ca(OH)2 (solubility rules).

Is Q > Ksp for Ca(OH)2?

[Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M

Ksp = [Ca2+][OH-]2 = 4.7 x 10-6

Q = [Ca2+]0[OH-]02 = 0.100 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form

If this solution also contained 0.100 M AlCl3, Aluminum could be precipitated,

with Calcium remaining in the solution.

터

1) Use the difference between Ksp’s to separate one

precipitate at a time.

2) Use Common Ion effect to Precipitate one Ion at a

time


터

The Common-Ion Effect and Solubility 01The Common-Ion Effect and Solubility 01

터

Factors That Affect SolubilityFactors That Affect Solubility

Mg2+(aq) + 2F1-(aq)MgF2(s)

Solubility and the Common-Ion Effect

터

1. Use the difference between Ksp’s to separate one


2. Use Common Ion effect to Precipitate one Ion at a time

3. Use pH to Increase or decrease solubility and

separate one ion at a time


터


Solubility and the pH of the Solution

Ca2+(aq) + HCO31-(aq) + H2O(l)CaCO3(s) + H3O1+(aq)

터

1. Use the difference between Ksp’s to separate one


2. Use Common Ion effect to Precipitate one Ion at a time

3. Use pH to Increase or decrease solubility and separate

one ion at a time.

4. Use complex formation to prevent certain Ions to

precipitate


터


Solubility and the Formation of Complex Ions

Ag(NH3)21+(aq)Ag(NH3)1+(aq) + NH3(aq)

Ag(NH3)21+(aq)Ag1+(aq) + 2NH3(aq) Kf = 1.7 x 107

K1 = 2.1 x 103Ag(NH3)1+(aq)Ag1+(aq) + NH3(aq)

K2 = 8.1 x 103

Kf is the formation constant.

How is it calculated?

(at 25 °C)

ѫ



= (2.1 x 103)(8.1 x 103) = 1.7 x 107[Ag(NH3)2

1+]

[Ag1+][NH3]2

K1 =[Ag(NH3)1+]

[Ag1+][NH3]K2 =

[Ag(NH3)21+]

[Ag(NH3)1+][NH3]

Kf = K1K2 =



Ag(NH3)21+(aq)Ag1+(aq) + 2NH3(aq)

터


Solubility and Amphoterism

Al(OH)41-(aq)Al(OH)3(s) + OH1-(aq)

Al3+(aq) + 6H2O(l)Al(OH)3(s) + 3H3O1+(aq)

In base:

In acid:

Aluminum hydroxide is soluble both in strongly acidic and in strongly basic solutions.

터


Solubility and Amphoterism

터

The Common Ion Effect and Solubility

The presence of a common ion decreases

the solubility of the salt.

What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr?

AgBr (s) Ag+ (aq) + Br- (aq)

Ksp = 5.4 x 10-13

s2 = Ksp

s = 7.3 x 10-7

NaBr (s) Na+ (aq) + Br- (aq)

[Br-] = 0.0010 M


[Ag+] = s

[Br-] = 0.0010 + s ≈ 0.0010

Ksp = 0.0010 x s

s = 5.4 x 10-10


M

M

터

• The presence of a common ion decreases the solubility.

• Insoluble bases dissolve in acidic solutions

• Insoluble acids dissolve in basic solution.

pH and SolubilitypH and Solubility

터

pH and Solubility

Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)

Ksp = [Mg2+][OH-]2 = 5.6 x 10-12

Ksp = (s)(2s)2 = 4s3

4s3 = 5.6 x 10-12

s = 1.1 x 10-4 M

[OH-] = 2s = 2.2 x 10-4 M

pOH = 3.65 pH = 10.35

At pH less than 10.35

Lower [OH-], Increase solubility of Mg(OH)2

OH- (aq) + H+ (aq) H2O (l)

remove

At pH greater than 10.35

Raise [OH-]

add

Decrease solubility of Mg(OH)2

What is the pH of a solution containing Mg(OH)2? what pH

Would reduce the solubility of this precipitate?

터

Complex Ion Equilibria and Solubility

A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.

Co2+ (aq) + 4Cl- (aq) CoCl4 (aq)2-

Kf =[CoCl4 ]

[Co2+][Cl-]4

2-

The formation constant or stability constant (Kf ) is the equilibrium constant for the complex ion formation.

Co(H2O)62+ CoCl4

2-

Kf

stability of complex

터

--

터

Qualitative AnalysisQualitative Analysis

터

Qualitative Analysis of

Cations

터








辠Ѯ

Flame Test for Cations

lithium sodium potassium copper

Nessler's is K2[HgI4] in a basic solution , [HgI4]2- reacts with the NH4+ to form

a yellow brown precipitate .

ѫ


ѫ

Separation of Ions by Selective PrecipitationSeparation of Ions by Selective Precipitation

• MS(s) + 2 H3O+(aq) ae M2+(aq) + H2S(aq) + 2 H2O(l)

• (see page 631) : K spa = [Cd+2 ] [H2S ] / [H3O+ ]2

• Kspa = For ZnS, Kspa = 3 x 10-2;• for CdS, Kspa = 8 x 10-7

• [Cd+2 ] = [Zn+2] = 0.005 M• Because the two cation concentrations are equal,• Qspa is :[Cd+2 ] [H2S ] / [H3O+ ]2 =(0.005).(0.1)/ (0.3)2

• Qspa = 6 x 10-3

• For CdS , Qspa > Kspa; CdS will precipitate. • For ZnS , Qspa< Kspa; Zn+2 will remain in solution.

Determine whether Cd+2 can be separated from Zn+2 by bubbling

H2S through ([H2S ] = 0.1 ) a 0.3 M HCl solutions that contains

0.005 M Cd+2 and 0.005 M Zn+2?

犰

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (Ka= 1.67 X 10-4 ).

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

Common ion effect

0.30 – x ≈ 0.30

0.52 + x ≈ 0.52

Mixture of weak acid and conjugate base!

X (0.52 + X)0.30 - X

= 1.67 X 10-4

X = 9.77 X 10-5 = [ H3O+]

pH = - log ( 9.77 X 10-5) = 4.01

춐

EndEnd

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