2023 NJC H2 Chemistry Prelim Paper 4 Ans

1 NJC SH2 Preliminary Examination 9729/04/23 [Turn over NATIONAL JUNIOR COLLEGE SH 2 Year − End Practical Examination Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 4 Practical Candidates answer on the Question paper. 9729/04 Tuesday 15 August 2023 2 hours 30 minutes READ THESE INSTRUCTIONS FIRST Write your identification number and name. Give details of the practical shift and laboratory where appropriate, in the boxes provided. Write in 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 scientific 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 19 and 20. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 20 printed pages and 0 blank page. Shift Laboratory For Examiner’s use 1 / 13 2 / 17 3 / 13 4 / 12 Total / 55

2 NJC SH2 Preliminary Examination 9729/04/23 1 Determine the reacting mole ratio of reagent X with Fe2+ Iron(II) ions, Fe2+, is readily oxidized. Fe2+ → Fe3+ + e– A common oxidising agent for Fe2+ ion is manganate(VII) ions in an acidic medium. MnO4 – + 8H+ + 5e–  Mn2+ + 4H2O Another oxidising agent that is also able to oxidise Fe2+ to Fe3+ .is reagent X. In this experiment, a limited amount of X is added to a solution of Fe2+. The resultant mixture containing unreacted Fe(II) ions will be titrated with acidified MnO4 –. FA 1 is a solution containing 0.075 mol dm–3 iron(II) sulfate, FeSO4. FA 2 is 0.010 mol dm–3 potassium manganate(VII), KMnO4. FA 3 is 0.025 mol dm–3 of a reagent X. FA 4 is 1.0 mol dm–3 sulfuric acid, H2SO4. (a) Procedure 1. Fill a burette with FA 3 and another burette with FA 2. 2. Pipette 25.0 cm 3 of FA 1 into a conical flask and add 10 cm 3 of FA 4 using a 10.0 cm 3 measuring cylinder into the same conical flask. 3. Run 12.00 cm 3 of FA 3 from the burette into the conical flask. 4. Titrate the mixture against FA 2 until the first permanent pink colour remains in the solution. You only need to do the titration ONCE. Record your titration results in the space below. Final burette reading / cm3 Initial burette reading / cm3 Volume of FA 2 added / cm3 24.50 (shift 1) 24.20 (shift 2) 24.00 (shift 3) Tables with correct headings and units(Do no award if did not put “burette”) (Do not award if any final and initial burette readings are inverted / if 50.00 is used as initial burette reading / burette reading is > 50.00) 1 All burette readings recorded to the nearest 0.05 cm 3 + correct computation of titres 1 Accuracy marks + 0.5 cm3 from teacher’s reading 1

3 NJC SH2 Preliminary Examination 9729/04/23 [Turn over [3] (b) (i) Calculate the amount of Fe2+ ions in FA 1 that reacted with MnO4 − in FA 2. Amount of MnO4 − used = 24.50 1000 × 0.01 = 2.45  10−4 mol Amount of Fe2+ reacted with MnO4 − = 2.45  10−4  5 = 1.23  10−3 amount of Fe2+ ions in FA 1 reacted with MnO4 − in FA 2 =……………………. mol [1] (ii) Calculate the total amount of Fe 2+ ions in FA 1 that was pipetted into the conical flask initially. Hence, calculate the amount of Fe2+ ions in FA 1 that reacted with X in FA 3 added. Total amount of Fe2+ = 25 1000 × 0.075 = 1.875  10−3 = 1.88  10−3 mol Total amount of Fe2+ ions in FA 1 =……………………. mol Total amount of Fe2+ = Fe2+ reacted with X + Fe2+ reacted with MnO4 − Amount of Fe2+ reacted with X = Total amount of Fe2+ − Fe2+ reacted with MnO4 − = 1.875  10−3 − 1.23  10−3 = 6.45  10−4 amount of Fe2+ ions in FA 1 reacted with X in FA 3 =……………………. mol [2] (iii) Calculate the amount of X in the 12.00 cm3 of FA 3 added. Hence, calculate the number of moles of Fe2+ ions that react with 1 mol of X. Amount of X used = 12 1000 × 0.025 = 3.00  10−4 mol amount of X =……………………………. mol amount of 𝐹𝑒2+ amount of 𝐗 = 𝟔.𝟒𝟓  𝟏𝟎−𝟒 3.00  10−4 = 2.15 number of moles of Fe2+ ions that react with 1 mol of X =……………………..[2] 1.23  10−3 6.45  10−4 3.00  10−4 2.15

4 NJC SH2 Preliminary Examination 9729/04/23 (iv) The redox half equations for the possible identity of X are given below. VO2+ + 2H+ + e– ⇌ V3+ + H2O H2O2 + 2H+ + 2e– ⇌ 2H2O NO3 − + 4 H+ + 3 e− ⇌ NO + 2H2O IO3 − + 5 H++ 4e− ⇌ HIO + 2H2O NO3 – + 10H+ + 8e– ⇌ NH4 + + 3H2O Using your answer from (b)(iii), deduce with reasoning, a possible identity for X. [You may use number of moles of Fe 2+ ions that react with 1 mol of X = 8.12 for this question if you did not get an answer for (b)(iii). Note that this value is not the correct answer for (b)(iii).] ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………...…………………………….[3] H2O2 Fe2+ is oxidized to form Fe3+ and thus X must be reduced. 1 mol of Fe2+ gives out 1 mol of electrons to form 1 mol of Fe3+. The reacting mole ratio for Fe2+ to X is approximately 2:1, indicating that every mol of X gains 2 mol of electrons. If using the value given (i.e. 8.12), NO3 – Fe2+ is oxidized to form Fe3+ and thus X must be reduced. 1 mol of Fe2+ gives out 1 mol of electrons to form 1 mol of Fe3+. The reacting mole ratio for Fe2+ to X is 8.12 :1, indicating that every mol of X gains 8 mol of electrons.

5 NJC SH2 Preliminary Examination 9729/04/23 [Turn over (v) Use the information below to calculate the percentage error for each volume measurement in Table 1.1. A 25 cm3 measuring cylinder is graduated to 1 cm3. The maximum error for a 25 cm3 pipette is ±0.06 cm3. Table 1.1 Solution Apparatus used Volume measured /cm3 % Error FA 1 Pipette 25.0 0.06 25  100% = 0.240% FA 2 Burette 23.50 2  0.05 23.50  100% = 0.426% FA 4 25 cm3 Measuring cylinder 10.0 0.5 10  100% = 5.00% [2] - Correctly calculate all 3 values Award 1 - must have  signs and - must have unit of % - all calculated values to 3 sf Award 1 For burette and measuring cylinder, the uncertainty from measure is half of the smallest division. Uncertainty for measuring cylinder = 0.5 cm3 Uncertainty for burette = 0.05 cm3 To measure a volume using a burette, two readings are required, hence total uncertainty for measurement = 2  (0.05 cm3) [Total: 13]

6 NJC SH2 Preliminary Examination 9729/04/23 2 Determine the order of reaction with respect to iodide In acidic solutions, iron(III) ions are reduced by iodide ions to form iron( II) ions. The iodide ions are oxidised to iodine. 2Fe3+(aq) + 2I−(aq)  2Fe2+(aq) + I2(aq) The iodine, I2, produced can be reacted immediately with thiosulfate ions, S2O3 2–. I2(aq) + 2S2O3 2‒(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 for the blue-black colour to appear. FA 5 is 0.0500 mol dm‒3 potass