| Date | November 2011 | Marks available | 1 | Reference code | 11N.2.hl.TZ0.4 |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Determine | Question number | 4 | Adapted from | N/A |
Question
Hypochlorous acid, HOCl(aq), is an example of a weak acid.
A household bleach contains sodium hypochlorite, NaOCl(aq), at a concentration of \({\text{0.705 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\). The hypochlorite ion, \({\text{OC}}{{\text{l}}^ - }{\text{(aq)}}\) is a weak base.
\[{\text{OC}}{{\text{l}}^ - }{\text{(aq)}} + {{\text{H}}_2}{\text{O(l)}} \rightleftharpoons {\text{HOCl(aq)}} + {\text{O}}{{\text{H}}^ - }{\text{(aq)}}\]
State the expression for the ionic product constant of water, \({K_{\text{w}}}\).
The \({\text{p}}{K_{\text{a}}}\) value of HOCl(aq) is 7.52. Determine the \({K_{\text{b}}}\) value of \({\text{OC}}{{\text{l}}^ - }{\text{(aq)}}\) assuming a temperature of 298 K.
Determine the concentration of \({\text{O}}{{\text{H}}^ - }{\text{(aq)}}\), in \({\text{mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\), at equilibrium and state one assumption made in arriving at your answer other than a temperature of 298 K.
Calculate the pH of the bleach.
Markscheme
\({\text{(}}{K_{\text{w}}} = {\text{)[}}{{\text{H}}^ + }{\text{(aq)][O}}{{\text{H}}^ - }{\text{(aq)]}}\);
Do not penalize if (aq) not stated.
H3O+ may be given instead of H+.
Do not mark awarded if square brackets are omitted or are incorrect.
\({\text{(p}}{K_{\text{b}}} = (14.00 - 7.52 = ){\text{ }}6.48{\text{ and) }}{K_{\text{b}}} = ({10^{ - 6.48}}) = 3.3 \times {10^{ - 7}}\);
Do not award mark if answer just left as 10–6.48.
\({K_{\text{b}}} = \frac{{{\text{[HOCl][O}}{{\text{H}}^ - }{\text{]}}}}{{{\text{[OC}}{{\text{l}}^ - }{\text{]}}}} = \frac{{{x^2}}}{{0.705}} = 3.3 \times {10^{ - 7}}\);
\({\text{[O}}{{\text{H}}^ - }{\text{]}} = 4.8 \times {10^{ - 4}}{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{)}}\);
Award [2] for correct value of [OH–].
\({\text{OC}}{{\text{l}}^ - }\) only partially hydrolysed / x negligible (compared to \({\text{OC}}{{\text{l}}^ - }\)) / OWTTE;
Accept [HOCl] = [OH–].
\({\text{[}}{{\text{H}}_3}{{\text{O}}^ + }{\text{]}}/{\text{[}}{{\text{H}}^ + }{\text{]}} = \frac{{{K_{\text{w}}}}}{{{\text{[O}}{{\text{H}}^ - }{\text{]}}}} = \frac{{1.00 \times {{10}^{ - 14}}}}{{4.8 \times {{10}^{ - 4}}}} = 2.1 \times {10^{ - 11}}\);
\({\text{pH}} = \left( { - {{\log }_{10}}{\text{[}}{{\text{H}}_3}{{\text{O}}^ + }{\text{]}}/ - {{\log }_{10}}{\text{[}}{{\text{H}}^ + }{\text{]}} = - {{\log }_{10}}(2.1 \times {{10}^{ - 11}}) = } \right)10.68\);
Award [2] for correct final answer.
Examiners report
This was generally well answered.
Many of the better students scored full marks here, and even the weaker students gained some marks.
Many of the better students scored full marks here, and even the weaker students gained some marks.
Many of the better students scored full marks here, and even the weaker students gained some marks.
