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
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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
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3. 10.35 cm3 of 0.105 mol dm-3 HCl was added to 25.0 cm3 of 0.170 mol dm-3 barium hydroxide solution. Calculate the pH of the resulting solution at 30°C, given that at this temperature Kw = 1.47 x 10-14 mol2 dm-6.
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4. Sodium hydroxide is a common ingredient in drain cleaner. 1.26g of drain cleaner was dissolved in water to form 100 cm3 of solution. The pH of this solution was 13.48. Determine the percentage by mass of sodium hydroxide in the drain cleaner.
Answers
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