| Date | May 2009 | Marks available | 1 | Reference code | 09M.2.hl.TZ1.4 |
| Level | HL | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate | Question number | 4 | Adapted from | N/A |
Question
The percentage of iron(II) ions, \({\text{F}}{{\text{e}}^{2 + }}\), in a vitamin tablet can be estimated by dissolving the tablet in dilute sulfuric acid and titrating with standard potassium manganate(VII) solution, \({\text{KMn}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\). During the process iron(II) is oxidized to iron(III) and the manganate(VII) ion is reduced to the manganese(II) ion, \({\text{M}}{{\text{n}}^{2 + }}{\text{(aq)}}\). It was found that one tablet with a mass of 1.43 g required \({\text{11.6 c}}{{\text{m}}^{\text{3}}}\) of \(2.00 \times {10^{ - 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) \({\text{KMn}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\) to reach the end-point.
State the half-equation for the oxidation of the iron(II) ions.
State the half-equation for the reduction of the \({\text{MnO}}_4^ - \) ions in acidic solution.
Deduce the overall redox equation for the reaction.
Calculate the amount, in moles, of \({\text{MnO}}_4^ - \) ions present in \({\text{11.6 c}}{{\text{m}}^{\text{3}}}\) of \(2.00 \times {10^{ - 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) \({\text{KMn}}{{\text{O}}_{\text{4}}}{\text{(aq)}}\).
Calculate the amount, in moles, of \({\text{F}}{{\text{e}}^{2 + }}\) ions present in the vitamin tablet.
Determine the percentage by mass of \({\text{F}}{{\text{e}}^{2 + }}\) ions present in the vitamin tablet.
Markscheme
\({\text{F}}{{\text{e}}^{2 + }} \to {\text{F}}{{\text{e}}^{3 + }} + {{\text{e}}^ - }\);
\({\text{MnO}}_4^ - + {\text{8}}{{\text{H}}^ + } + {\text{5}}{{\text{e}}^ - } \to {\text{M}}{{\text{n}}^{2 + }} + {\text{4}}{{\text{H}}_2}{\text{O}}\);
\({\text{MnO}}_4^ - + {\text{5F}}{{\text{e}}^{2 + }} + {\text{8}}{{\text{H}}^ + } \to {\text{M}}{{\text{n}}^{2 + }} + {\text{5F}}{{\text{e}}^{3 + }} + {\text{4}}{{\text{H}}_2}{\text{O}}\);
Accept e instead of e–.
\({\text{amount of MnO}}_4^ - = \frac{{11.6}}{{1000}} \times 0.0200 = 2.32 \times {10^{ - 4}}{\text{ mol}}\);
\({\text{amount of F}}{{\text{e}}^{2 + }} = 5 \times 2.32 \times {10^{ - 4}} = 1.16 \times {10^{ - 3}}{\text{ mol}}\);
\({\text{mass of F}}{{\text{e}}^{2 + }} = 55.85 \times 1.16 \times {10^{ - 3}} = 6.48 \times {10^{ - 2}}{\text{ g}}\);
\({\text{percentage of F}}{{\text{e}}^{2 + }}{\text{ in tablet}} = \frac{{6.48 \times {{10}^{ - 2}}}}{{1.43}} = 100 = 4.53\% \);
Examiners report
This question was generally well answered. A common mistake with writing half-equations was the failure to realise that only single arrows should be used if oxidation and reduction are specifically asked for. Candidates were only penalized once for this error.
Given that the half-equation involving \({\text{MnO}}_4^ - \) ions is provided in the Data Booklet, it was surprising that several candidates could not correctly write the equation for their reduction in acidic solution.
