State an equation for the reaction of ethanoic acid with water.

Calculate the pH of \({\text{0.200 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ethanoic acid \(({\text{p}}{K_{\text{a}}} = 4.76)\).

Determine the pH of a solution formed from adding \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) ethanoic acid, \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH(aq)}}\), to \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.600 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) sodium hydroxide, NaOH(aq).

(if acid added) \({\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}^ + } \to {\text{C}}{{\text{H}}_3}{\text{COOH}}\);

(if alkali added) \({\text{C}}{{\text{H}}_3}{\text{COOH}} + {\text{O}}{{\text{H}}^ - } \to {\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}_2}{\text{O}}\);

Explanation marks cannot be awarded without equations.

\({\text{C}}{{\text{H}}_3}{\text{COOH(aq)}} + {{\text{H}}_2}{\text{O(l)}} \rightleftharpoons {\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }{\text{(aq)}} + {{\text{H}}_3}{{\text{O}}^ + }{\text{(aq)}}\);

OR

\({\text{C}}{{\text{H}}_3}{\text{COOH(l)}} + {{\text{H}}_2}{\text{O(l)}} \rightleftharpoons {\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }{\text{(aq)}} + {{\text{H}}_3}{{\text{O}}^ + }{\text{(aq)}}\);

OR

\({\text{C}}{{\text{H}}_3}{\text{COOH(aq)}} \rightleftharpoons {\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }{\text{(aq)}} + {{\text{H}}^ + }{\text{(aq)}}\);

Must include \( \rightleftharpoons \).

Ignore state symbols.

(ii)     \({K_{\text{a}}} = {10^{ - 4.76}}/1.74 \times {10^{ - 5}}/{\text{pH}} = {\text{p}}{K_{\text{a}}} + \log \frac{{{\text{[SALT]}}}}{{{\text{[ACID]}}}}\);

\(1.74 \times {10^{ - 5}} = \frac{{{{{\text{[}}{{\text{H}}^ + }{\text{]}}}^2}}}{{{\text{0.200}}}}/{\text{[}}{{\text{H}}^ + }{\text{]}} = 0.00187\);

\({\text{pH}} = 2.73\);

Award [3] for correct final answer, allow mark for correct conversion of [H⁺] to pH even if [H⁺] incorrect.

(initial) \({\text{[C}}{{\text{H}}_{\text{3}}}{\text{COOH]}} = 0.500{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) and) eqm \({\text{[C}}{{\text{H}}_{\text{3}}}{\text{COOH]}} = 0.200{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\);

(initial) \({\text{[C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }{\text{]}} = 0.300{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) and) eqm \({\text{[C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }{\text{]}} = 0.300{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\);

Allow 0.02 moles and 0.03 moles instead of 0.200 and 0.300.

\({\text{[}}{{\text{H}}^ + }{\text{]}} = {K_{\text{a}}}\frac{{{\text{[C}}{{\text{H}}_3}{\text{COOH]}}}}{{{\text{[C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }{\text{]}}}} = 1.16 \times {10^{ - 5}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\);

\({\text{pH}} = 4.94\);

Award [3 max] for correct final answer if no working shown.

The only issue was that some candidates forgot the reversible arrow in the equation.

A pleasing number were able to complete the pH calculation successfully.

Only the best candidates scored full marks for the buffer calculation; in some cases an incorrect expression was used, but more often there was no attempt to calculate the equilibrium amounts or concentrations.

There were very few who could write appropriate equations for the buffer action, even though it clearly stated that the answer should include equations many explained buffer action without any equations and scored no marks as a result.