2024 H2 Chem P1 Q16 — Acid base equilibrium Solution

The answer is D (statement 2 only is correct).
Statement 1 is incorrect.
\(n(\text{OH}^-)\) added \(= \frac{50.0}{1000} \times 0.100 = 0.00500\) mol
\(n(\text{CH}_3\text{CO}_2\text{H}) = \frac{25.0}{1000} \times 0.150 = 0.00375\) mol
Excess \(\text{OH}^- = 0.00500 - 0.00375 = 0.00125\) mol in a total volume of 75.0 \(\text{cm}^3\)
\([\text{OH}^-] = \frac{0.00125}{0.0750} = 0.0167\) mol \(\text{dm}^{-3}\), pOH \(= -\lg 0.0167 = 1.78\), pH \(= 14.00 - 1.78 = 12.2\)
A pH of 13.0 is the pH of the 0.100 mol \(\text{dm}^{-3}\) NaOH itself. At the end, some \(\text{OH}^-\) has been used up by the acid and the solution has been diluted, so the pH is 12.2, not 13.0.
Statement 2 is correct.
When half of the acid has been neutralised, the amounts of \(\text{CH}_3\text{CO}_2^-\) and \(\text{CH}_3\text{CO}_2\text{H}\) are equal, so \(\frac{[\text{salt}]}{[\text{acid}]} = 1\).
pH \(= \text{p}K_a + \lg\frac{[\text{salt}]}{[\text{acid}]} = \text{p}K_a + \lg 1 = \text{p}K_a\)
• A and C are wrong because statement 1 is incorrect.
• B is wrong because statement 2 is correct.