Calculate the \({K_{\text{a}}}\) value of benzoic acid, \({{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\), using Table 15 in the Data Booklet.

Based on its \({K_{\text{a}}}\) value, state and explain whether benzoic acid is a strong or weak acid.

Determine the hydrogen ion concentration and the pH of a \({\text{0.010 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) benzoic acid solution. State one assumption made in your calculation.

\({K_{\text{a}}} = 6.310 \times {10^{ - 5}}/6.31 \times {10^{ - 5}}\);

Accept 6.3 \( \times \) 10^–5.

weak (acid);

\({K_{\text{a}}} \ll 1/{\text{small }}{K_{\text{a}}}\);

\({\text{[}}{{\text{H}}_3}{{\text{O}}^ + }{\text{]}}/{\text{[}}{{\text{H}}^ + }{\text{]}} = \sqrt {{K_{\text{a}}} \times 0.010} \);

\({\text{[}}{{\text{H}}_3}{{\text{O}}^ + }{\text{]}}/{\text{[}}{{\text{H}}^ + }{\text{]}} = 7.9 \times {10^{ - 4}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);

\({\text{pH}} = 3.10/3.1/3.12\);

Award [3] for correct final answer of pH.

assume \({\text{x}} \ll 0.010{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\) / ionization of water is insignificant / \({{\text{[}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH]}}_{{\text{initial}}}} = {{\text{[}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH]}}_{{\text{eq}}}}\) / temperature 25 °C /298 K;

Part (a) (i) proved to be a well known topic where only weaker candidates couldn’t finish the calculation.

For (ii) although a significant number of candidates knew that benzoic acid was a weak acid, only the better candidates explained this based on the fact that \({K_{\text{a}}}\) is \( \ll 1\).

Part (iii) was very well answered, but even the better candidates often forgot to state one assumption made in the calculation.