2022 DHS Y6 H2 Prelim Paper 4 Suggested Solutions

This document consists of 16 printed pages. © DHS 2022 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 25 August 2022 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 21 and 22. 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 10 3 14 4 15 Total 55

2 © DHS 2022 9729/04 Answer all questions in the spaces provided. 1 Determination of solubility product, Ksp, of magnesium carbonate The solubility of solid magnesium carbonate, MgCO3, in water is low. In this experiment, you will determine the solubility product, Ksp, of magnesium carbonate using volumetric analysis. MgCO3(s) ⇌ Mg2+(aq) + CO32–(aq) A saturated solution of magnesium carbonate was prepared by collecting the filtrate after mixing aqueous solutions of magnesium sulfate and sodium carbonate. You will then determine the amount of carbonate ions left in the filtrate using hydrochloric acid. FA 1 is a saturated solution of magnesium carbonate, MgCO3. FA 2 is 0.003 mol dm–3 hydrochloric acid, HCl. methyl orange indicator Note: • Read the procedures given in part (a) and part (b)(i). • Part (a) has been done for you. You are not required to carry out part (a), but you will still need to read part (a) to know how FA 1 is prepared. (a) Preparation of FA 1 1. Use a measuring cylinder to transfer 20 cm 3 of 0.100 mol dm–3 magnesium sulfate to a clean and dry 250 cm3 beaker. 2. Use a measuring cylinder to transfer 50 cm3 of 0.0400 mol dm–3 sodium carbonate to the same beaker. A white precipitate forms. 3. Stir the mixture thoroughly. Leave this mixture to stand for 15 minutes to allow equilibrium to be reached. 4. Filter the reaction mixture into a dry 250 cm3 conical flask, using dry filter funnel and filter paper. The filtrate is labelled as FA 1. Do not wash the white precipitate with water. (b) (i) Titration of FA 1 against FA 2 1. Fill a burette with FA 2. 2. Pipette 10.0 cm3 of FA 1 into a 100 cm3 conical flask. 3. Add a few drops of methyl orange indicator to the conical flask. 4. Titrate the solution in the conical flask with FA 2. The end-point for this titration is reached when the solution changes colour from yellow to orange. 5. Record all burette readings, to an appropriate level of precision, in the space provided on page 3. 6. Repeat steps 1 to 5 until consistent results are obtained.

3 © DHS 2022 9729/04 [Turn over Titration results Final burette reading / cm3 17.00 34.00 Initial burette reading / cm3 0.00 17.00 Volume of FA 2 used / cm3 17.00 17.00 [3] (ii) From your titrations, obtain a suitable volume of FA 2 to be used in your calculations. Show clearly how you obtained this volume. Average volume of FA 2 used = 17.00 + 17.00 2 = 17.00 cm3 volume of FA 2 = ………………………….cm3 [4] (c) (i) The equation for the reaction in the titration is shown below. 2H+(aq) + CO32–(aq) → H2O(l) + CO2(g) Calculate the concentration of CO32– ions in FA 1. Moles of HCl reacted = ( 17.00 1000) (0.003) = 5.1 x 10–5 mol 2H+ ≡ CO32– Moles of CO32– in 10.0 cm3 of FA 1 = ( 1 2) (5.1 ×10-5) = 2.55 x 10–5 mol Concentration of CO32– = 2.55 × 10-5 0.01 = 2.55 x 10–3 mol dm–3 concentration of CO32– ions = …………………………. mol dm–3 [1] (ii) Calculate the total amount of CO32– ions present in the total volume of filtrate prepared in (a). Total moles of CO32– in 70.0 cm3 of FA 1 = ( 70 10) (2.55 ×10-5) = 1.785 x 10–4 = 1.79 x 10–4 mol total amount of CO32– ions = …………………………. mol [1] (iii) Hence, calculate the amount of CO32– ions precipitated as MgCO3. Initial moles of CO32– = ( 50 1000) (0.04) = 2 x 10–3 mol Total moles of CO32– precipitated as MgCO3 = 2 x 10–3 – 1.785 x 10–4 = 1.8215 x 10–3 = 1.82 x 10–3 mol amount of CO32– ions precipitated as MgCO3 = …………………………. mol [2]

4 © DHS 2022 9729/04 (iv) Deduce the amount of Mg2+ ions removed by precipitation, in step 3 of the procedure in (a). Hence, calculate the amount of Mg2+ ions left in FA 1. MgCO3(s) ⇌ Mg2+(aq) + CO32–(aq) Moles of Mg2+ removed by precipitation = 1.82 x 10–3 mol Initial moles of Mg2+ = ( 20 1000) (0.1) = 2 x 10–3 mol Moles of Mg2+ left in FA 1 = 2 x 10–3 – 1.8215 x 10–3 = 1.785 x 10–4 = 1.79 x 10–4 mol amount of Mg2+ ions removed by precipitation = …………………………. mol amount of Mg2+ ions left = …………………………. mol [2] (v) Write an expression for the solubility product, Ksp, of magnesium carbonate. Include units in your answer. Ksp of MgCO3 = [Mg2+][CO32–] mol2 dm−6 [1] (vi) Calculate a value for the solubility product, Ksp, of magnesium carbonate. Ksp of MgCO3 = ( 1.785×10-4 70 1000 ) ( 1.785×10-4 70 1000 ) = 6.50 x 10–6 mol2 dm−6 Ksp = ………………………………... [1] (d) A student follows the procedures described and obtained a higher Ksp value compared to the literature value of 6.82 × 10–6 at 25 °C. Give a possible explanation for the higher Ksp value obtained. [1] The student did not carry out the experiment at 25 °C. He probably carried out the experiment at a temperature at which the equilibrium position of the reaction lies more towards the Mg2+ and CO32– ions / the forward reaction is favoured. Hence, the Ksp value obtained was higher. [Total: 16]

5 © DHS 2022 9729/04 [Turn over 2 Planning Avogadro’s constant, L, is defined as the number of particles in one mole of a substance. The currently accepted value is 6.02 × 1023 mol–1. The Avogadro’s constant, L, can be determined through the electrolysis of acidified aqueous potassium iodide. During electrolysis, the amount of material discharged at each of the electrodes depends solely on the amount of current that has passed through the system. A student conducted the electrolysis experiment. When a current was passed through a solution of acidified aqueous potassium iodide, hydrogen gas was produced at one electrode, while iodine was produced at the other electrode. The amount of hydrogen produced can be determined from the volume of hydrogen gas collected by downward displacement of water. The amount of iodine produced can be determined by titration of a portion of the resultant solution with aqueous sodium thiosulfate. I2 + 2S2O32– → 2I– + S4O62– (a) Explain why the volume of gas collected was higher than the theoretical volume of hydrogen gas that can be produced in the electrolysis. [1] The volume of the hydrogen gas that is measured using the downward displacement of water also includes the volume of water vapour. (b) (i) Plan an investigation to determine the amount of iodine produced in the electrolysis of acidified aqueous