At 25°C, 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is added respectively, are : Given : Ka = 5 10-4 pKa = 3.3 1
- array|c|c| (a) & (b) \ 0.7 & 2.0 \ array
- array|c|c| (a) & (b) \ 2.0 & 3.3 \ array
- array|c|c| (a) & (b) \ 1.1 & 2.2 \ array
- array|c|c| (a) & (b) \ 3.0 & 2.2 \ array
Answer
(2) array|c|c| (a) & (b) \ 2.0 & 3.3 \ array
Solution
(a) With the weak acid alone and 1,
[H+] = Ka C = 5 10-4 0.2 = 10-4 = 10-2 M, so pH = 2.0.
(b) HX = 20 × 0.2 = 4 mmol and NaOH = 10 × 0.2 = 2 mmol, so after reaction HX = 2 mmol and X- = 2 mmol; this is half-neutralisation, giving an acidic buffer, hence
pH = pKa + [X-][HX] = 3.3 + 1 = 3.3
Answer: (2)