Question
Chloroethene, C2H3Cl, is an important organic compound used to manufacture the polymer poly(chloroethene).
State an equation for the reaction of ethanoic acid with water.[1]
Calculate the pH of \({\text{0.200 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) ethanoic acid \(({\text{p}}{K_{\text{a}}} = 4.76)\).[3]
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).[4]
(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.[2]
Answer/Explanation
Markscheme
\({\text{C}}{{\text{H}}_3}{\text{COOH(aq)}} + {{\text{H}}_2}{\text{O(l)}} \rightleftharpoons {\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ – }{\text{(aq)}} + {{\text{H}}_3}{{\text{O}}^ + }{\text{(aq)}}\);
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.
[N/A]
Examiners report
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.
Question
A buffer solution is made using \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.500 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) nitric acid, \({\text{HN}}{{\text{O}}_{\text{3}}}{\text{(aq)}}\), and \({\text{25.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{1.00 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) ammonia solution, \({\text{N}}{{\text{H}}_{\text{3}}}{\text{(aq)}}\).
Bromocresol green is an acid–base indicator. Information about bromocresol green is given in Table 16 of the Data Booklet.
A solution of ammonia has a concentration of \({\text{0.500 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\).
Calculate the pH of the ammonia solution using information from Table 15 of the Data Booklet. State one assumption made.[4]
(i) State the meaning of the term buffer solution.
(ii) Calculate the concentrations of ammonia and ammonium ion in the buffer solution.
(iii) Determine the pH of the buffer solution at 25 °C.
(iv) Explain why the pH of the buffer solution is different from the pH of the ammonia solution calculated in (d).
(v) Explain the action of the buffer solution when a few drops of nitric acid solution are added to it.[8]
(i) Identify the property of bromocresol green that makes it suitable to use as an acid–base indicator.
(ii) State and explain the relationship between the pH range of bromocresol green and its \({\text{p}}{K_{\text{a}}}\) value.[3]
Answer/Explanation
Markscheme
\({\text{[O}}{{\text{H}}^ – }{\text{]}} = \left( {\sqrt {0.500 \times 1.78 \times {{10}^{ – 5}}} } \right) = 2.98 \times {10^{ – 3}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\);
\({\text{pOH}} = – {\log _{10}}{\text{[O}}{{\text{H}}^ – }{\text{]}} = 2.526/{\text{[}}{{\text{H}}^ + }{\text{]}} = \left( {\frac{{1.00 \times {{10}^{ – 14}}}}{{2.98 \times {{10}^{ – 3}}}}} \right) = 3.35 \times {10^{ – 12}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\);
\({\text{pH}} = 11.47\);
Accept correct answer obtained using another method.
Assumption:
\({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}} = 0.500{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) / \({\text{[NH}}_4^ + {\text{]}} = {\text{[O}}{{\text{H}}^ – }{\text{]}}\) / all \({\text{O}}{{\text{H}}^ – }\) ions come from the reaction of ammonia with water and not from the dissociation of water / temperature is 25 °C/298 K / OWTTE;
(i) resists change in pH when small amounts of a strong base/strong acid/water are added to it;
(ii) \({\text{[N}}{{\text{H}}_{\text{3}}}{\text{]}} = 0.250{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\);
\({\text{[NH}}_4^ + {\text{]}} = 0.250{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\);
(iii) \({\text{pOH}} = {\text{p}}{K_b} = 4.75\);
\({\text{pH}} = 9.25\);
(iv) equilibrium shifted left in buffer / OWTTE;
(v) acid neutralized by hydroxide / most of the added \({{\text{H}}^ + }\) ions react with \({\text{N}}{{\text{H}}_{\text{3}}}\);
more ammonia reacts with water to replace hydroxide ions / more \({\text{NH}}_4^ + \) ions form so there is little change in the pH / OWTTE;
Accept equations.
(i) colours of HIn and \({\text{I}}{{\text{n}}^ – }\) are different / OWTTE;
(ii) colour change occurs when \({\text{[HIn]}} = {\text{[I}}{{\text{n}}^ – }{\text{]}}\);
\({\text{pH}} = {\text{p}}{K_{\text{a}}}\);
OR
pH range is a range of pH values either side of \({\text{p}}{K_{\text{a}}}\) value;
lower pH when acid colour is seen and upper pH when alkaline colour seen;
Examiners report
Calculation of pH in (d) proved challenging for some and straightforward for others. Those who knew how to perform the calculations generally also correctly stated an assumption.
Most candidates correctly described a buffer solution in (e). Several candidates had difficulty calculating the concentrations of ammonia and ammonium ions in the buffer but managed to calculate the pH correctly (some with ECF). The explanations of why the pH of the buffer differs from the pH of ammonia and the action of the buffer when a few drops of nitric acid are added were poorly done and would have been better with the use of equations and references to equilibrium.
Answers to (f) were quite general. Many candidates simply said that bromocresol green changes colour with no further details, or said that the indicator had different colours in acid and alkaline conditions. Most candidates scored 1 mark for stating that the \({\text{p}}{K_{\text{a}}}\) is in the middle of the pH range.