John E. McMurry • Robert C. Fay

C H E M I S T R Y

Chapter 15Chapter 15Applications of Aqueous EquilibriaApplications of Aqueous Equilibria

Neutralization ReactionNeutralization Reaction General Formula

Acid + Base - Water + Salt

Neutralization ReactionsNeutralization Reactions

H2O(l)H1+(aq) + OH1-(aq)

Assuming complete dissociation:

Strong Acid-Strong Base

After neutralization: pH = 7(net ionic equation)

H2O(l) +NaOH(aq)HCl(aq) + NaCl(aq)

2H2O(l)H3O1+(aq) + OH1-(aq)or

Strong acid-Strong base Strong acid-Strong base neutralizationneutralization

When the number moles of acid and base are mixed together[H3O+] = [-OH] = 1.0 x 10-7M

Reaction proceeds far to the right

Kn =1

[H3O+] [-OH]

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

Neutralization ReactionsNeutralization Reactions

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

Assuming complete dissociation:

Weak Acid-Strong Base

After neutralization: pH > 7(net ionic equation)

H2O(l) +CH3CO2H(aq) + NaOH(aq) NaCH3CO2(aq)

Weak acid-strong base Weak acid-strong base neutralizationneutralization

Neutralization of any weak acid by a strong base goes 100% to completion

-OH has a great infinity for protons

Kn =1

KaKw

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

Neutralization ReactionsNeutralization Reactions

NH41+(aq)H1+(aq) + NH3(aq)

Assuming complete dissociation:

Strong Acid-Weak Base

After neutralization: pH < 7

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

orH2O(l) +H3O1+(aq) + NH3 (aq) NH4

1+(aq)(net ionic equation)

Strong acid-weak base Strong acid-weak base neutralizationneutralization

Neutralization of any weak base by a strong acid goes 100% to completion

H3O+ has a great infinity for protons

Kn =1

KbKw

H2O(l) +H3O1+(aq) + NH3 (aq) NH41+(aq)

Neutralization ReactionsNeutralization Reactions

NH41+(aq) + CH3CO2

1-(aq)CH3CO2H(aq) + NH3(aq)

Weak Acid-Weak Base

After neutralization: pH = ?

NH4CH3CO2(aq)CH3CO2H(aq) + NH3(aq)

(net ionic equation)

Weak acid-weak base Weak acid-weak base neutralizationneutralization Less tendency to proceed to completion than

neutralization involving strong acids and strong bases

Kn =1

KbKw

Ka

ExamplesExamples Write balanced net ionic equations for the

neutralization of equal molar amounts of the following acids and bases. Indicate whether the pH after the neutralization is greater than, equal to or less than 7

HNO2 and KOH Ka HNO2 = 4.5 x 10-4

The Common-Ion EffectThe Common-Ion Effect

H3O+(aq) + CH3CO2–(aq)CH3CO2H(aq) + H2O(l)

Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium

The Common-Ion EffectThe Common-Ion EffectThe pH of 0.10 M acetic acid is 2.89. Calculate the pH of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution.

The Common-Ion EffectThe Common-Ion Effect

The Common-Ion EffectThe Common-Ion Effect

H3O+(aq) + CH3CO2–(aq)CH3CO2H(aq) + H2O(l)

The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left.

Le Châtelier’s Principle

The Common-Ion EffectThe Common-Ion Effect

ExampleExample In 0.15 M NH3, the pH is 11.21 and the percent

dissociation is 1.1%. Calculate the concentrations of NH3, pH and percent dissociation of ammonia in a solution that is 0.15M and 0.45 MNH4Cl

Buffer SolutionsBuffer SolutionsBuffer Solution: A solution which contains a weak acid and its conjugate base and resists drastic changes in pH.

CH3CO2H + CH3CO21-

HF + F1-

NH41+ + NH3

H2PO42- + HPO4

2-

Weak acid

+

Conjugate base

ForExample:

Buffer SolutionsBuffer Solutions Add a small amount of base (-OH) to a buffer solution

◦ Acid component of solution neutralizes the added base

Add a small amount of acid (H3O+) to a buffer solution◦ Base component of solution neutralizes the added acid

The addition of –OH or H3O+ to a buffer solution will change the pH of the solution, but not as drastically as the addition of –OH or H3O+ to a non-buffered solution

Buffer SolutionsBuffer Solutions

Conjugate base(M+A-)

Weak acid

H3O1+(aq) + A1-(aq)HA(aq) + H2O(l)

Addition of OH1- to a buffer:

H2O(l) + A1-(aq)HA(aq) + OH1-(aq)100%

Addition of H3O1+ to a buffer:

H2O(l) + HA(aq)A1-(aq) + H3O1+(aq)100%

Buffer SolutionsBuffer Solutions

ExampleExample pH of human blood (pH = 7.4) controlled by conjugated

acid-base pairs (H2CO3/HCO3-). Write the neutralization

equation fro the following effects◦ With addition of HCl

◦ With addition of NaOH

ExampleExample Calculate the pH of 0.100L of a buffer solution that is

0.25M in HF and 0.50 M in NaF. ◦ What is the change in pH on addition of 0.002 mol

HCl◦ What is the change in pH on addition of 0.010

moles KOH◦ Calculate the pH after addition of 0.080 moles HBr* Assume the volume remains constant

ExampleExample Calculate the pH of the buffer that results from mixing

60.0mL of 0.250M HCHO2 and 15.0 mL of 0.500M NaCHO2

Ka = 1.8 x 10-4

◦ Calculate the pH after addition of 10.0 mL of 0.150 MHBr. Assume volume is additive

The Henderson-Hasselbalch The Henderson-Hasselbalch EquationEquation

Conjugate baseWeak acid

H3O1+(aq) + Base(aq)Acid(aq) + H2O(l)

[Acid]

[Base][H3O1+] = Ka

Ka =[H3O1+][Base]

[Acid]

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

The Henderson-Hasselbalch The Henderson-Hasselbalch EquationEquation

[Base]

[Acid]pH = pKa + log

[Acid]

[Base][H3O1+] = Ka

-log([H3O1+]) = -log(Ka) - log[Acid]

[Base]

ExamplesExamples Calculate the pH of a buffer solution that is 0.50 M in

benzoic acid (HC7H5O2) and 0.150 M in sodium benzoate (NaC7H5O2). Ka = 6.5 x 10-5

How would you prepare a NaHCO3-Na2CO3 buffer solution that has pH = 10.40 Ka2 = 4.7 x 10-11