A-LevelChemistryEquilibria9701/43 · May/June 2021 · Q87 Marks

The sketch graph for the titration of ethanoic acid, CH3CO₂H, with sodium hydroxide is shown. [Figure titration curve]

(a)The sketch graph for the titration of ethanoic acid, CH3CO₂H, with sodium hydroxide is shown. [Figure titration curve]
(a)(i)In the region circled on the graph, identify the two organic species that are present in the solution. Explain why the pH of the mixture only changes slowly and gradually in this region when sodium hydroxide is being added.[3]
(a)(ii)The equivalence point in this acid-base titration is where the two solutions have been mixed in exactly equal molar proportion. Suggest why the pH is greater than 7 at the equivalence point in this titration.[1]
(b)An impure sample of ammonium vanadate(V), NH4VO3, with mass 0.150g, is dissolved in an excess of dilute acid. In this solution all vanadium is present as VO2+ ions. An excess of zinc powder is added to the solution and all the VO2+ ions are reduced to V2+ ions. The mixture is filtered to remove any remaining zinc powder. VO2+ + 4H+ + 3e¯ → V2+ + 2H2O When the resulting solution is titrated, 20.10 cm³ of 0.0250 mol dm³ acidified MnO4¯ oxidises all V2+ ions back to VO2+ ions. MnO4 + 8H+ + 5e¯ → Mn2+ + 4H2O Calculate the percentage by mass of NH4VO3 in the 0.150g impure sample of NH4VO3. Give your answer to three significant figures. [M: NH4VO3, 116.9] percentage by mass of NH4VO3 =[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[3] M1: CH3CO2H and CH3CO2-
[3] M2: due to buffering action / acting as a buffer solution
[3] M3: CH3CO2H reacts with NaOH / OH- (forming CH3CO2- and water) OR OH- reacts with H+ and equilibrium CH3CO2H ⇌ CH3CO2- + H+ shifts to the right
[1] identifying CH3CO2¯ is present (with water) at the equivalence point OR CH3CO2¯ react with water forming OH- OR titrating a weak acid with a strong base
[3] M1: moles MnO4 = 0.025 × 0.0201 = 5.025 × 10-4 moles V2+ = 5.025 × 10-4 × 5/3 = 8.375 × 10-4
[3] M2: moles VO3¯ = 8.375 × 10-4 mass of NH4VO3 = 116.9 × 8.375 × 10–4 = 0.0979 g
[3] M3: % Purity of NH4VO3 = 100 × 0.0979 / 0.15 = 65.3 must be 3 sf

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