Chapter 15
Applications of
Aqueous Equilibria
15.1 Neutralization Reaction

General Formula
Acid + Base  Water + Salt
Four types:
Strong acid – strong base
Weak acid – strong base
Strong acid – Weak base
Weak acid-Weak base
To what extend does a neutralization
Reaction go to completion?
Strong Acid-Strong Base

If we mix equal numbers moles of HCl(aq) and NaOH(aq), the
[H3O+] and [-OH] remaining in solution after neutralization will
be same as those in pure water
HCl(aq)
+
NaOH(aq)
H2O(l) +
Assuming complete dissociation:
(net ionic equation)
H3O1+(aq) + OH1-(aq)
[H3O+] = [-OH] = 1.0 x 10-7M
Reaction proceeds far to the right
pH = 7.00
Kn = 1.0 x 1014
2H2O(l)
NaCl(aq)
Weak Acid-Strong Base

Because a weak acid HA is largely undissociated, the net equation
for the neutralization reaction of a weak acid with a strong base
involves proton transfer from HA to the strong base, -OH
CH3CO2H(aq)
+
NaOH(aq)
(net ionic equation) CH3CO2H(aq) + OH1-(aq)




H2O(l)
+
NaCH3CO2(aq)
H2O(l) + CH3CO21-(aq)
Neutralization of any weak acid by a strong base goes 100% to
completion
-OH has a great infinity for protons
pH >7.00
Kn > 107
Strong Acid-Weak Base

A strong acid HA is completely dissociated into H3O+ and A- ions,
and its neutralization reaction with a weak base involves proton
transfer from H3O+ to the weak base B:
HCl(aq)
(net ionic equation)
+ NH3(aq)
H3O1+(aq) + NH3 (aq)
NH4Cl(aq)
H2O(l) + NH41+(aq)
Neutralization of any weak base by a strong acid goes 100% to
completion
 H3O+ has a great infinity for protons
 pH < 7.00
 Kn > 107
Weak Acid-Weak Base
CH3CO2H(aq) + NH3(aq)
NH4CH3CO2(aq)
(net ionic equation)
CH3CO2H(aq) + NH3(aq)
NH41+(aq) + CH3CO21-(aq)
After neutralization: pH = ?
Weak acid-weak base neutralization has less tendency to proceed to completion
than strong acids and strong bases.
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)
What are the ions present in the solution?
15.2 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. Ka = 1.8 x 10-5
Step 1: Identify the mixture compositions
Step 2: Consider all possible reactions
Step 3: Identify the principle reaction
Step 4: Calculate new pH
Step 5 : Check your answer
The Common-Ion Effect
Le Châtelier’s Principle
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.
H3O+(aq) + CH3CO2–(aq)
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 NH3 and 0.45 M NH4Cl
Kb = 1.8 x 10-5

15.3 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
H2PO41- + 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
Buffer Solutions
Example

pH of human blood (pH = 7.4) controlled by conjugated acid-base
pairs (H2CO3/HCO3-). Write a neutralization equation for 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
• Ka = 3.5 x 10-4
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 = 1.7 x 10-4
◦ Calculate the pH after addition of 10.0 mL of 0.150 MHBr.
Assume volume is additive

Buffer Capacity




A measure of amount of acid or base that the solution can
absorb without a significant change in pH.
Depends on how many moles of weak acid and conjugated base
are present.
For an equal volume of solution: the more concentrated the
solution, the greater buffer capacity
For an equal concentration: the greater the volume, the
greater the buffer capacity
Example

The following pictures represent solutions that contain a weak
acid HA and/ or its sodium salt NaA. (Na+ ions and solvent water
molecules have been omitted for clarity

Which of the solutions are buffer solution?
Which solution has the greatest buffer capacity?

15.4 The Henderson-Hasselbalch
Equation
Weak acid
Conjugate base
Acid(aq) + H2O(l)
Ka =
[H3O1+][Base]
[H3O1+] = Ka
[Acid]
pH = pKa + log
H3O1+(aq) + Base(aq)
[Base]
[Acid]
[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