Chapter 15
Applications of
Aqueous Equilibria
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?
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)
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-
Example
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. Is this a buffer solution?
Ka = 1.8 x 10-5

Example

Calculate the pH of 0.15M HF and 0.25NaF mixture. Is this a
buffer solution?
Example

Calculate the pH of 100.0 mL DI water

Calculate the new pH after adding 1.0 mL of 0.10M HCl to the
above water solution.
Buffer Solutions
HA(aq) + H2O(l)
H3O1+(aq) + A1-(aq)
Conjugate base
(M+A-)
Add a small amount of base (-OH) to a buffer solution
◦ Acid component of solution neutralizes the added base
Weak acid

Addition of OH1- to a buffer:
HA(aq) +
OH1-(aq)
100%
H2O(l) + A1-(aq)
Buffer Solutions
HA(aq) + H2O(l)
H3O1+(aq) + A1-(aq)
Conjugate base
(M+A-)
Weak acid

Add a small amount of acid (H3O+) to a buffer solution
◦ Base component of solution neutralizes the added acid
Addition of H3O1+ to a buffer:
A1-(aq)
+
H3O1+(aq)
100%
H2O(l) + HA(aq)
Buffer Solutions

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 H O+ to a non-buffered solution
3
Buffer Solutions
Example

pH of human blood (pH = 7.4) controlled by conjugated acid-base
pairs (H2CO3/HCO3-). Write an equation for this buffer mixture then
neutralization equation for the following effects
◦ With addition of HCl
◦ With addition of NaOH
Example

50.0 mL of 0.100 M HCl was added to a .100L buffer consisting
of 0.025 moles of sodium acetate and 0.030 moles of acetic acid.
What is the pH of the buffer before and after the addition of the
acid? Ka of acetic acid is 1.7 x 10-5. Assume the volume is
constant
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 0.010 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?

Example

What is the maximum amount of acid that can be added to a
buffer made by the mixing of 0.35 moles of sodium hydrogen
carbonate with 0.50 moles of sodium carbonate? How much
base can be added before the pH will begin to show a
significant change?
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
Example

How would you prepare a NaHCO3-Na2CO3 buffer solution that
has pH = 10.40 Ka2 = 4.7 x 10-11
Example

You prepare a buffer solution of .323 M NH3 and (NH4)2SO4.
What molarity of (NH4)2SO4 is necessary to have a pH of 8.6?
(pKb NH3= 4.74)