Describe the composition of an acidic buffer solution.

Determine the pH of a buffer solution, correct to two decimal places, showing your working, consisting of 10.0 g of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ and 10.0 g of CH$_3$COONa in ${\text{0.250 d}}{{\text{m}}^{\text{3}}}$ of solution. ${K_{\text{a}}}$ for ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} = 1.8 \times {10^{ - 5}}$ at 298 K.

(solution containing significant/equal amounts of a) weak acid and its salt / (solution containing) strong base to which excess of weak acid has been added / OWTTE;

Accept (solution containing) weak acid and conjugate base.

Do not accept descriptions with specific compounds alone (e.g. CH₃COOH and CH₃COONa) unless compounds are stated as weak acid and its salt.

Accept answer such as (solution containing) x mol of weak acid and $\frac{1}{2}x\,mol$ of strong base.

${M_{\text{r}}}{\text{(C}}{{\text{H}}_{\text{3}}}{\text{COOH)}} = 60.06$ and ${M_{\text{r}}}{\text{ C}}{{\text{H}}_3}{\text{COONa}} = 82.04$;

${\text{[C}}{{\text{H}}_3}{\text{COOH]}} = 6.66 \times {10^{ - 1}}/0.666{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$

${\text{[C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }{\text{]}} = 4.88 \times {10^{ - 1}}/{\text{ }}0.488{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$;

${\text{[}}{{\text{H}}_3}{{\text{O}}^ + }{\text{]}}/{\text{[}}{{\text{H}}^ + }{\text{]}} = (1.8 \times {10^{ - 5}} \times 6.66 \times {10^{ - 1}})/4.88 \times {10^{ - 1}} = 2.46 \times {10^{ - 5}}/0.0000246{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$;

${\text{pH}} = \left( { - {\text{log[}}{{\text{H}}_3}{{\text{O}}^ + }{\text{]}} =  - {\text{log(2.46}} \times {\text{1}}{{\text{0}}^{ - 5}}{\text{)}} = } \right){\text{ 4.61 (2dp)}}$;

Award [5] for correct final answer of pH = 4.61 with some working shown.

Award [2 max] for pH = 4.61 without any working at all shown.

Two decimal places are required for M5.

OR

${M_{\text{r}}}{\text{(C}}{{\text{H}}_{\text{3}}}{\text{COOH)}} = 60.06$ and ${M_{\text{r}}}{\text{ C}}{{\text{H}}_3}{\text{COONa}} = 82.04$;

${\text{[C}}{{\text{H}}_3}{\text{COOH]}} = 6.66 \times {10^{ - 1}}/0.666{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$

${\text{[C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }{\text{]}} = 4.88 \times {10^{ - 1}}/{\text{ }}0.488{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$;

${\text{pH}} =  - {\text{log(1.8}} \times {\text{1}}{{\text{0}}^{ - 5}}{\text{)}} + {\text{log}}\frac{{{\text{[salt]}}}}{{{\text{[acid]}}}}$;

$= \left( {4.74 + \log \frac{{0.488}}{{0.666}} = 4.74 - 0.135 = } \right){\text{ }}4.61{\text{ (2dp)}}$;

M4 can be scored even if not explicitly stated if M5 is correct based on previous values.

Award [5] for correct final answer of pH = 4.61 with some working shown.

Award [2 max] for pH = 4.61 without any working at all shown.

Two decimal places are required for M5.

This question was based on buffer solutions and was found to be quite challenging for candidates. In part (a), some candidates again failed to read the question, which asked for a description of an acidic buffer solution. Many did not state explicitly that a weak acid is involved (acid alone was not sufficient).

In part (b), only the best candidates scored all five marks. In addition to conceptual errors, there were also a number of transcription errors (molar mass and arithmetic errors). Candidates also were required to express their answer to two decimal places. A number of candidates used the Henderson-Hasselbalch equation, but often an incorrect equation was given.