2023 DHS H2 Chem Prelim Paper 4 Suggested Solutions

© DHS 2023 9729/04 [Turn over Suggested Solutions DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY Paper 4 Practical Candidates answer on the Question Paper. 9729/04 24 August 2023 2 hours 30 minutes READ THESE INSTRUCTIONS FIRST Write your centre number, index number, name and class at the top of this page. Give details of the practical shift and laboratory where appropriate, in the boxes provided. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved sci entific calculator is expected, where appropriate. You may lose marks if you do not show your working or if you do not use appropriate units. Qualitative Analysis Notes are printed on pages 23 and 24. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Shift Laboratory For Examiner’s Use 1 16 2 13 3 15 4 11 Total 55

2 © DHS 2023 9729/04 Answer all questions in the spaces provided. 1 Determination of change in oxidation number in XO3– oxidant. FA 1 is 0.0265 mol dm–3 potassium manganate(VII), KMnO4(aq) FA 2 is 0.0750 mol dm–3 XO3– oxidant solution. X is the symbol of an element whose identity is not needed in this question. FA 3 is a solution containing 0.150 mol dm–3 iron(II) ions, Fe2+. FA 3 solution is prepared by dissolving an iron(II) salt in sulfuric acid. In the presence of acid, XO3– oxidises iron(II) to iron(III) while itself is reduced to X b+. In order to determine b, a sample of XO3– is added to a known amount of iron(II) in excess. The amount of remaining iron(II) ions can then be determined by titration against manganate(VII) ions as shown in equation 1. equation 1 MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) (a) Procedure 1. Pipette 25.0 cm3 of FA 3 into a conical flask. 2. Using another pipette, add exactly 10.0 cm3 of FA 2 to FA 3 in the conical flask. 3. Fill the burette with FA 1. 4. Titrate the mixture in the conical flask with FA 1 until the appearance of the first permanent pale orange or pale pink colour. 5. Repeat the titration as many times as you think necessary to obtain accurate results. 6. Record your results of titration in a suitable tabulated form below. Results 1 2 Final burette reading / cm3 17.00 34.00 Initial burette reading / cm3 0.00 17.00 Volume of FA 1 (KMnO4) / cm3 17.00 17.00 [2] (b) From your titrations, obtain a suitable volume of FA 1, VFA 1, to be used in your calculations. Show clearly how you have obtained this volume. VFA 1 = 17.00+17.00 2 = 17.00 cm3 (2dp) VFA 1 = ……………….. cm3 [3] (c) (i) Calculate the amount of Fe2+ ions pipetted into the conical flask. Moles/amount of Fe2+ = 25.0 1000 𝑥 0.150 ✓ ✓

3 © DHS 2023 9729/04 [Turn over = 0.00375 mol amount of Fe2+ ions = …0.00375 mol…… [1] (ii) Calculate the amount of XO3– ions present in the conical flask. Moles/amount of XO3– = 10.0 1000 𝑥 0.075 = 0.000750 mol amount of XO3– ions = …0.000750 mol ……. [1] (iii) Using your answer to (b), calculate the amount of Fe2+ ions that remain after reaction with XO3–. Number of moles of KMnO4 = 17.00 1000 𝑥 0.0265 = 4.505 x 10–4 mol Since MnO4– ≡ 5Fe2+, Number of moles of Fe2+ = 5 x 4.505 x 10–4 = 2.2525 x 10–3 mol amount of Fe2+ ions remaining = …2.25 x 10–3 mol ……. [1] (iv) Calculate the amount of Fe2+ ions that reacted with XO3– ions in the conical flask and hence, the amount of Fe2+ that reacted with one mole of XO3– ions. Number of moles of Fe2+ that reacted with XO3– = 0.00375 – 0.0022525 = 1.4975 x 10–3 Number of moles of Fe2+ that reacted with one mole of XO3– ions = 0.0014975 0.00075 = 2.00 amount of Fe2+ that reacted with one mole of XO3– ions = …2.00 mol… [2] (v) Hence, calculate the final oxidation number of the element X when Fe2+ reacts with XO3–. Initial oxidation number of X in 𝑋𝑂3 − = +5 Fe2+ → Fe3+ + e– Since 2Fe2+ ≡ 𝑋𝑂3 −, total mol of electrons gained per mol of 𝑋𝑂3 − = 2 final oxidation number of X in 𝑋𝑂3 − = 5 – 2 = +3

4 © DHS 2023 9729/04 final oxidation number of the element X = …+3 ……. [2] (d) A student carries out the same experiment in (a). However unknowingly, he uses a sample of FA 2 with concentration 0.100 mol dm–3 instead. Suggest what effect this will have on the volume of FA 1 obtained. Explain your answer. The amount of FA 2 increases and hence the amount of FA 3 after reacting with FA 2 decreases. Less amount of FA 1 is required to react with the remaining FA 3 and the volume of FA 1 will be smaller. [1] (e) (i) A student followed the procedure described in (a) using a different oxidant instead of FA 2. The student obtained a mean titre value of 22.20 cm 3. The actual volume of FA 1 required should have been 22.40 cm3. Calculate the difference between the actual value and the student's value as a percentage. This is the experimental error. experimental error = (│22.20 – 22.40│/ 22.40) x 100% = 0.893% experimental error = ……….……………..% [1] (ii) Using his mean titre value, the student calculates that the maximum total percentage error of the apparatus used is 0.77%. Use the following data to verify the student’s calculation. apparatus uncertainty 25 cm3 pipette 0.03 cm3 10 cm3 pipette 0.02 cm3 Hence, determine if the student performed the experiment well. Explain your answer. maximum total percentage error = ( 0.03 25 + 0.02 10 + 0.10 22.20) × 100% = 0.770 % Since the experimental error (0.893%) is bigger than the maximum total percentage error of the apparatus used (0.770%), the student did not perform the experiment well. OR

5 © DHS 2023 9729/04 [Turn over Since the difference between the experimental error (0.893%) and the maximum total percentage error of the apparatus used (0.770%) is very small (0.123%), the student did perform the experiment well. [2] [Total: 16]

6 © DHS 2023 9729/04 2 Determination of the kinetics of the reaction between Fe3+ ions and iodide ions, I−. You are provided with the following reagents. FA 4 contains 0.0250 mol dm–3 acidified iron(III) chloride, FeCl3 FA 5 is 0.0800 mol dm−3 aqueous potassium iodide, KI FA 6 is 0.00500 mol dm−3 sodium thiosulfate, Na2S2O3 starch solution Fe3+ ions oxidise iodide ions, I–, to iodine, I2, as shown in equation 2. In this experiment, you will investigate how the rate of this reaction is affected by the concentration of Fe3+ ions. equation 2 2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq) A fixed and small amount of thiosulfate ions, S2O32–, and starch indicator will be added to a mixture of Fe3+(aq) and I–(aq). The iodine, I2, produced reacts immediately with thiosulfate ions, S 2O32−, as shown in equation 3. equation 3 I2(aq) + 2S2O32−(aq) → 2I−(aq) + S4O62−(aq) When all the thiosulfate has been used, the iodine produced will turn starch indicator blue-black. The rate of the reaction can therefore be measured by finding the time it takes for the reaction mixture to turn blue–black. You will perform a series of four experiments. In the series of experiments, the rate equation for the reaction can be simpli