Once again, we can assume that strong bases are fully dissociated in solution so the concentration of OH– is the same as the original concentration of the base.
NaOH (aq) ⇾ Na+ (aq) + OH– (aq)
To find the pH of a base, we need to use the expression for the ionic product of water, Kw.
Kw = [H+(aq)] [OH–(aq)] = 1.00 x 10-14 mol2 dm-6
If we know the value of Kw and the [OH–(aq)], then we can find [H+(aq)] and hence the pH.
E.g. what is the pH of 0.1 mol dm-3 NaOH?
Kw = [H+(aq)] [OH–(aq)]
1.00 x 10-14 = [H+(aq)] . 0.1
[H+(aq)] = 1 x 10-14 / 0.1 = 1.00 x 10-13
pH = -log [H+(aq)] = 13
Weak bases are only partially dissociated in solution so we cannot assume that the concentration of the base equals [OH–(aq)] . In order to find the [OH–(aq)] we need to know where the equilibrium position lies.
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH– (aq)
Kb for NH3 = 1.8 x 10-5 mol dm-3 so the equilibrium position lies strongly to the left.
E.g. find the pH of a 0.15 mol dm-3 solution of NH3 given the the pKb of ammonia is 4.75.
[OH–(aq)]2 = 2.67 x 10-6
[OH–(aq)] = 0.00163 mol dm-3
Kw = [H+(aq)] [OH–(aq)]
1.00 x 10-14 = [H+(aq)] . 0.00163
[H+(aq)] = 6.132 x 10-12 mol dm-3
pH = – log [H+(aq)] = 11.2
Practice questions
- Calculate the pH of the following solutions of strong bases:
(a) 0.15 mol dm-3 LiOH
(b) 0.25 mol dm-3 Mg(OH)2
2. Calculate the pH of the following solutions of weak bases:
(a) 0.033 mol dm-3 CH3NH2, Kb = 5.6 x 10-4 mol2 dm-6
(b) 0.68 mol dm-3 CH3COO–, Kb = 5.6 x 10-10 mol2 dm-6
Answers
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