2024 H2 Chem P1 Q17 — Acid base equilibrium Solution

The answer is C (4.93).
\(n(\text{KOH}) = \frac{10.0}{1000} \times 0.20 = 0.00200\) mol
\(n(\text{C}_2\text{H}_5\text{CO}_2\text{H}) = \frac{25.0}{1000} \times 0.15 = 0.00375\) mol
\(\text{C}_2\text{H}_5\text{CO}_2\text{H} + \text{OH}^- \rightarrow \text{C}_2\text{H}_5\text{CO}_2^- + \text{H}_2\text{O}\)
The acid is in excess, so \(n(\text{C}_2\text{H}_5\text{CO}_2^-)\) formed \(= 0.00200\) mol and \(n(\text{C}_2\text{H}_5\text{CO}_2\text{H})\) left \(= 0.00375 - 0.00200 = 0.00175\) mol.
Both are in the same total volume (35.0 \(\text{cm}^3\)), so the ratio of amounts equals the ratio of concentrations.
\(\text{p}K_a = -\lg(1.35 \times 10^{-5}) = 4.870\)
pH \(= \text{p}K_a + \lg\frac{[\text{salt}]}{[\text{acid}]} = 4.870 + \lg\frac{0.00200}{0.00175} = 4.870 + 0.058 = 4.93\)
• A is wrong: 4.60 comes from using the initial 0.00375 mol of acid, \(4.870 + \lg\frac{0.00200}{0.00375}\), without subtracting the acid used up.
• B is wrong: 4.81 comes from inverting the ratio, \(4.870 + \lg\frac{0.00175}{0.00200}\).
• D is wrong: 5.14 comes from inverting the ratio and using the initial acid, \(4.870 + \lg\frac{0.00375}{0.00200}\).