John E. McMurry • Robert C. Fay
C H E M I S T R Y
Chapter 15
Applications of Aqueous Equilibria
Neutralization Reaction

General Formula
Acid + Base - Water + Salt
Neutralization Reactions
Strong Acid-Strong Base
HCl(aq)
+
NaOH(aq)
H2O(l) +
NaCl(aq)
Assuming complete dissociation:
H1+(aq) + OH1-(aq)
H2O(l) or
H3O1+(aq) + OH1-(aq)
(net ionic equation)
After neutralization: pH = 7
2H2O(l)
Strong acid-Strong base neutralization
H3O1+(aq) + OH1-(aq)


2H2O(l)
When the number moles of acid and base are mixed together
[H3O+] = [-OH] = 1.0 x 10-7M
Reaction proceeds far to the right
Neutralization Reactions
Weak Acid-Strong Base
CH3CO2H(aq)
+
NaOH(aq)
H2O(l)
+
NaCH3CO2(aq)
Assuming complete dissociation:
CH3CO2H(aq) + OH1-(aq)
H2O(l) + CH3CO21-(aq)
(net ionic equation)
After neutralization: pH > 7
Weak acid-strong base neutralization


Neutralization of any weak acid by a strong base goes 100% to completion
-OH has a great infinity for protons
CH3CO2H(aq) + OH1-(aq)
H2O(l) + CH3CO21-(aq)
Neutralization Reactions
Strong Acid-Weak Base
HCl(aq)
+ NH3(aq)
NH4Cl(aq)
Assuming complete dissociation:
or
H1+(aq) + NH3(aq)
H3O1+(aq)
+ NH3 (aq)
NH41+(aq)
H2O(l) +
After neutralization: pH < 7
NH41+(aq)
(net ionic equation)
Strong acid-weak base neutralization


Neutralization of any weak base by a strong acid goes 100% to
completion
H3O+ has a great infinity for protons
H3O1+(aq) + NH3 (aq)
H2O(l) + NH41+(aq)
Neutralization Reactions
Weak Acid-Weak Base
CH3CO2H(aq) + NH3(aq)
NH4CH3CO2(aq)
CH3CO2H(aq) + NH3(aq)
NH41+(aq) + CH3CO21-(aq)
(net ionic equation)
After neutralization: pH = ?
Weak acid-weak base neutralization

Less tendency to proceed to completion than neutralization
involving strong acids and strong bases
Examples


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 Effect
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
CH3CO2H(aq) + H2O(l)
H3O+(aq) + CH3CO2–(aq)
The Common-Ion Effect
The 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 Effect
The Common-Ion Effect
Le Châtelier’s Principle
CH3CO2H(aq) + H2O(l)
H3O+(aq) + CH3CO2–(aq)
The addition of acetate ion to a solution of acetic acid suppresses the
dissociation of the acid. The equilibrium shifts to the left.
The Common-Ion Effect
Example

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 Solutions
Buffer Solution: A solution which contains a weak acid and
its conjugate base and resists drastic changes in pH.
Weak acid
+
Conjugate base
For
Example:
CH3CO2H + CH3CO21HF + F1NH41+ + NH3
H2PO42- + HPO42-
Buffer 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 Solutions
HA(aq) + H2O(l)
H3O1+(aq) + A1-(aq)
Weak acid
Conjugate base
(M+A-)
Addition of OH1- to a buffer:
HA(aq) +
OH1-(aq)
100%
H2O(l) + A1-(aq)
Addition of H3O1+ to a buffer:
A1-(aq)
+
H3O1+(aq)
100%
H2O(l) + HA(aq)
Buffer Solutions
Example

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
Example

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
Example
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 Equation
CH3CO2H(aq) + H2O(l)
Weak acid
Conjugate base
Acid(aq) + H2O(l)
Ka =
H3O1+(aq) + CH3CO21-(aq)
[H3O1+][Base]
[Acid]
H3O1+(aq) + Base(aq)
[H3O1+] = Ka
[Acid]
[Base]
The Henderson-Hasselbalch Equation
[H3O1+]
= Ka
[Acid]
[Base]
[Acid]
-log([H3O1+]) = -log(Ka) - log
pH = pKa + log
[Base]
[Acid]
[Base]
Examples

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