2017 H2 Chem P1 Q13 — Acid base equilibrium Solution

The answer is D \(\left(\dfrac{997}{18}L\sqrt{K_\text{c}}\right)\).
Concentration of water: \([\text{H}_2\text{O}] = \dfrac{0.997 \text{ g cm}^{-3}}{18 \text{ g mol}^{-1}} = \dfrac{0.997}{18} \text{ mol cm}^{-3} = \dfrac{997}{18} \text{ mol dm}^{-3}\)
In pure water \([\text{H}_3\text{O}^+] = [\text{OH}^-]\), so \(K_\text{c} = \dfrac{[\text{H}_3\text{O}^+]^2}{[\text{H}_2\text{O}]^2}\) and \([\text{H}_3\text{O}^+] = \sqrt{K_\text{c}}\,[\text{H}_2\text{O}] = \dfrac{997}{18}\sqrt{K_\text{c}} \text{ mol dm}^{-3}\).
In 1.00 \(\text{dm}^3\): \(n(\text{H}_3\text{O}^+) = \dfrac{997}{18}\sqrt{K_\text{c}}\) mol, so the number of ions \(= \dfrac{997}{18}L\sqrt{K_\text{c}}\), where \(L\) is the Avogadro constant.
• A is wrong: \(\dfrac{0.997}{18}\) is the concentration of water in mol \(\text{cm}^{-3}\) (it gives the number of ions in 1 \(\text{cm}^3\), not 1 \(\text{dm}^3\)), and it uses \(K_\text{c}\) instead of \(\sqrt{K_\text{c}}\).
• B is wrong: it uses the concentration of water in mol \(\text{cm}^{-3}\), so it gives the number of ions in 1 \(\text{cm}^3\), which is 1000 times too small.
• C is wrong: it uses \(K_\text{c}\) instead of \(\sqrt{K_\text{c}}\). \([\text{H}_3\text{O}^+]^2 = K_\text{c}[\text{H}_2\text{O}]^2\), so the square root must be taken.