1. Chemistry 1011 TOPIC TEXT REFERENCE Acids and Bases
Masterton and Hurley Chapter 4.2 (Review), 13, 14.1, 15.1 (page 427), 21.2 (page589)
Chemistry 1011 Slot 5

2. 14.1 Buffers YOU ARE EXPECTED TO BE ABLE TO:
Explain why a buffer solution is resistant to changes in pH.
Calculate the pH of a pure buffer solution.
Calculate the pH of a buffer solution to which a known quantity of a strong acid or base has been added.
Chemistry 1011 Slot 5

3. What is a Buffer Solution?
A buffer solution is able to maintain a stable pH when quantities of acid or base are added
If one drop (0.1mL) of 0.1mol/L HCl is added to 10mL pure water, the pH will change from 7 to 3
If one drop (0.1mL) of 0.1mol/L HCl is added to 10mL of a buffer solution, the pH will not change appreciably
Chemistry 1011 Slot 5

4. What is a Buffer Solution?
A buffer solution is an aqueous mixture of a weak acid or a weak base with its salt
An aqueous mixture of ammonia and ammonium chloride is a buffer solution
An aqueous mixture of acetic acid and sodium acetate is a buffer solution
In each case, small quantities of strong acid or strong base can be added without altering the pH
Chemistry 1011 Slot 5

5. The Acetic Acid/Acetate Ion Buffer
A solution is prepared that contains 1.0 x 10-1 mol/L acetic acid and 1.0 x 10-1 mol/L sodium acetate
CH3COOH(aq) CH3COO-(aq) + H+(aq)
CH3COONa(aq) CH3COO-(aq) + Na+(aq)
The sodium acetate is a salt and is fully ionized
The [CH3COO-] is 1.0 x 10-1 mol/L
(Ignore the acetate ion produced by the acetic acid. This concentration is << 1.0 x 10-1 mol/L )
The [CH3COOH] is 1.0 x 10-1 mol/L
(Ignore the acetic acid that has dissociated. This concentration is << 1.0 x 10-1 mol/L )
Chemistry 1011 Slot 5

6. The Acetic Acid/Acetate Ion Buffer
For acetic acid:
Ka = [CH3COO-]x[H+] = 1.8 x 10-5
[CH3COOH]
Ka = (1.0 x 10-1)x[H+] = 1.8 x 10-5
(1.0 x 10-1)
[H+] = 1.8 x 10-5 mol/L
pH = 4.7
Chemistry 1011 Slot 5

7. Example – pH of a Buffer Solution
Calculate the [OH-] and pH of a buffer solution prepared so that [CH3NH2] is 0.300mol/L and [CH3NH3+] is 0.400mol/L
CH3NH2 is a weak base and CH3NH3+ is its conjugate acid
CH3NH2 + H2O(aq) CH3NH3 +(aq) + OH-(aq)
Kb = [CH3NH3 +] x[OH-] = 4.2 x 10-4
[CH3NH2]
Kb = (4.00 x 10-1) x [OH-] = 4.2 x 10-4
(3.00 x 10-1)
[OH-] = (4.2 x 10-4) x (3.00 x 10-1) = 3.15 x 10-4
(4.00 x 10-1)
pH = 10.5
Chemistry 1011 Slot 5

8. Adding a Strong Acid to the Buffer
The buffer is a mixture of a weak acid and its conjugate base
CH3COOH(aq) + H2O(aq) CH3COO-(aq) + H3O+(aq)
The concentrations of undissociated acetic acid and acetate ions are quite large. (If they are equal, then pH = pKa)
The concentration of H+ (H3O+) is small
If a strong acid is added (adding more H+ ) the equilibrium system moves to remove the added H+
Some of the acetate ions react with the added H+ ions to produce undissociated acetic acid. The pH remains largely unaltered
Chemistry 1011 Slot 5

9. Adding a Strong Acid to the Buffer
10mL of 1.0 x 10-1mol/L HCl is added to 100mL of a buffer solution containing 1.0 x 10-1mol/L CH3COOH and 1.0 x 10-1mol/L CH3COO-
Initial moles CH3COOH = 1.0 x 10-2 mol
Initial moles CH3COO- = 1.0 x 10-2 mol
Added moles H+ = 1.0 x 10-3 mol
CH3COOH(aq) CH3COO-(aq) + H+(aq)
The added H+ ions react with CH3COO- ions to form CH3COOH
Chemistry 1011 Slot 5

10. Adding a Strong Acid to the Buffer
Resulting moles CH3COOH = 1.0 x x 10-3 mol = 1.1 x mol
Resulting moles CH3COO- = 1.0 x x 10-3 mol = 0.9 x mol
New [CH3COOH] = 1.1 x mol = 1.0 x 10-1mol/L
0.11L
New [CH3COO-] = 0.9 x mol = 8.0 x mol/L
Ka = [CH3COO-]x[H+] = 1.8 x 10-5
[CH3COOH]
Ka = (8.0 x 10-2 )x[H+] = 1.8 x 10-5
(1.0 x 10-1)
[H+] = 2.3 x 10-5 mol/L pH = 4.6 (was 4.7)
Chemistry 1011 Slot 5

11. Choosing a Buffer
Different systems require buffer solutions of different pH
The pH depends on
Ka or Kb for the weak acid or base (Greatest impact)
The ratio of amounts of weak acid or base and the salt
If the concentrations of weak acid or base and salt are equal, then pH will equal p Ka or pKb
Chemistry 1011 Slot 5

12. Effectiveness of Buffers
To be effective, there must be sufficient acid or base and salt to react with any added strong acid or base
Usually there will be similar amounts of weak acid or base and salt
This produces a buffer with a pH  pKa or pKb that is able to absorb either strong acid or strong base
Once so much strong acid or base has been added that all of either the weak acid or its conjugate base have reacted, then the buffer will not be effective
Chemistry 1011 Slot 5