2025 H2 Chem Prelim Paper 4 (Solutions)_ACJC

© ACJC 2025 9729/04/Prelim/2025 1 Determination of the stoichiometry of the reaction between hydroxylamine and iron(III) ion A redox reaction takes place between hydroxylamine, NH 2OH, and iron(III) ion, Fe 3+, in an acidic medium. The iron(III) ion is converted to iron(II) ion, Fe2+. The reaction is slow at room temperature, but is completed in a few minutes when heated to 90 oC. The iron(II) ions formed can then be oxidised by manganate(VII) ions for quantitative analysis. You are to determine, by titration, which of the following unbalanced equations represents the reaction between hydroxylamine and iron(III) ion. equation 1.1 NH2OH + Fe3+ → Fe2+ + …N2 + …H+ + …H2O equation 1.2 NH2OH + 2Fe3+ → 2Fe2+ + …N2O + …H+ + …H2O equation 1.3 NH2OH + 3Fe3+ → 3Fe2+ + …NO + …H+ FA 1 is 0.0150 mol dm−3 potassium manganate(VII), KMnO4 FA 2 is a solution prepared by boiling a 1.00 dm3 aqueous mixture of 3.30 g of hydroxylamine hydrochloride, NH2OH.HCl, excess iron(III) chloride, FeCl3, and excess sulfuric acid, H2SO4 FA 3 is 1.00 mol dm−3 sulfuric acid, H2SO4 (a) Procedure 1. Fill the burette with FA 1. 2. Using the pipette, transfer 25.0 cm3 of FA 2 into the conical flask. 3. Using a measuring cylinder, transfer 10.0 cm3 of FA 3 to the same conical flask. 4. Titrate the mixture in the conical flask with FA 1. The end-point is reached when the first permanent pale pink colour is seen. 5. Record your titration results, to an appropriate level of precision, in the space provided. 6. Repeat steps 2 to 5 to obtain consistent results. (i) Titration results Final burette reading / cm3 24.70 24.70 Initial burette reading / cm3 0.00 0.00 Volume of FA 1 used / cm3 24.70 24.70 [2]

© ACJC 2025 9729/04/Prelim/2025 [Turn over (ii) From your titration results in (b)(i), obtain a suitable volume of FA 1, VFA 1, to be used in your calculations. Show clearly how you obtained this volume. VFA1 = (24.70 + 24.70) / 2 = 24.70 cm3 VFA 1 = ....................................... [4] (b) (i) Calculate the amount of MnO4– present in VFA 1 obtained in (a)(ii). 24.70/1000 x 0.0150 = 0.0003705 = 3.71 x 10–4 mol (3 s.f.) amount of MnO4– = ....................................... [1] (ii) Calculate the amount of Fe2+ in 25.0 cm3 of FA 2. MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l) Fe2+(aq) → Fe3+(aq) + e- 5 x 0.0003705 = 0.0018525 mol = 0.00185 mol (3 s.f.) amount of Fe2+ = ....................................... [1] (iii) Calculate the amount of NH2OH.HCl that reacted in the FA 2 pipetted into the conical flask. [Ar: H, 1.0; N, 14.0; O, 16.0; Cl, 35.5] Other than the fact that the reaction to prepare FA 2 was complete and there were no impurities in the chemicals used, state an assumption made in your calculation. Mr of NH2OH.HCl = 69.5 Amount of NH2OH.HCl in 1.0 dm3 = 3.3 / 69.5 = 0.04748 mol Amount of NH2OH•HCl in 25.0