2018 H2 Chem P1 Q14 — Acid base equilibrium Solution

The answer is A.
Each water molecule that ionises gives one \(\text{H}^+\) and one \(\text{OH}^-\) (\(\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-\)), so in pure water \([\text{H}^+] = [\text{OH}^-]\) at every temperature, and \([\text{H}^+] = \sqrt{K_\text{w}}\).
At 10 °C: \([\text{H}^+] = \sqrt{0.293 \times 10^{-14}} = 5.41 \times 10^{-8}\text{ mol dm}^{-3}\), pH \(= -\lg(5.41 \times 10^{-8}) = 7.27\)
At 40 °C: \([\text{H}^+] = \sqrt{2.92 \times 10^{-14}} = 1.71 \times 10^{-7}\text{ mol dm}^{-3}\), pH \(= -\lg(1.71 \times 10^{-7}) = 6.77\)
\(K_\text{w}\) increases with temperature because the ionisation of water is endothermic. So \([\text{H}^+]\) rises and the pH decreases, while the water stays neutral.
• B is wrong: the pH does decrease, but \([\text{H}^+]\) cannot be greater than \([\text{OH}^-]\) in pure water; they are always formed in equal amounts.
• C is wrong: \([\text{H}^+]\) increases, so the pH decreases (7.27 to 6.77), not increases.
• D is wrong on both counts: the pH decreases, and \([\text{H}^+]\) always equals \([\text{OH}^-]\) in pure water.