HCI 2024 Prelim P4 Solutions

2024 HCI C2 H2 Chemistry Prelims / Paper 4 HWA CHONG INSTITUTION 2024 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 4 1 (a) (i) Test 2 Obs 1. white/ off-white ppt formed (with 1cm3 NaOH(aq)) Obs 2. brown ppt formed/ brown ppt observed on the wall of the boiling tube/ white ppt turned brown Obs 3. (white) ppt partially soluble / insoluble in excess NaOH Obs 4. brown residue (on filtering) Obs 5. colourless filtrate (on filtering) Obs 6. white ppt formed (when dilute H2SO4 is added to the filtrate) Obs 7. (white) ppt soluble in excess dilute H2SO4 Test 3 Obs 8. white ppt (with dropwise addition of NH3(aq)) Obs 9. brown ppt/ some white ppt turns brown/ brown ppt observed on the wall of the test-tube Obs 10. ppt insoluble in excess aq. ammonia Test 4 Obs 11. white ppt (with aq. barium nitrate) Obs 12. (white ppt) insoluble in dilute HNO3 There are 12 Obs. They are scaled to 6 marks as follows. 11 – 12 = [6] 9 – 10 = [5] 7 – 8 = [4] 5 – 6 = [3] 3 – 4 = [2] 1 – 2 = [1] 0 = [0] 1 (a) (ii) Al3+: Based on test 2, white ppt forms in the filtrate when dilute H2SO4 is added to it. [1] Mn2+: Based on test 2 (&/or 3), white ppt rapidly turns brown. [1]

1 SO42–: Based on test 4, white ppt insoluble in dilute HNO3 [1] 1 (a) (iii) Effervescence of CO2 is due to the acid-base reaction between CO32– and the H+(aq) produced from the hydrolysis/ionisation of [Al(H2O)6]3+. [1] The white ppt observed is due to the precipitation of insoluble MnCO3 and Al(OH)3. [1] 1 (b) (i) Obs 1. pale brown / colourless filtrate (after dilute HNO3 is added) Obs 2. solution decolourises / remains colourless (when FA 2 is added) Obs 3. brown ppt/solution observed when NaOH(aq) is first added Obs 4. brown ppt dissolves/ brown solution decolorises shortly after appearing Obs 5. this is followed by effervescence of colourless, odourless gas Obs 6. that relights a glowing splint Obs 7. gas evolved is oxygen Obs 8. brown ppt formed dissolves more slowly (with excess NaOH(aq)) / brown ppt remains eventually. (effervescence ceases) There are 8 Obs. They are scaled to 3 marks as follows. 6 – 8 = [3] 4 – 5 = [2] 2 – 3 = [1] 0 – 1 = [0] 1 (b) (ii) The brown intermediate formed on adding NaOH(aq) increases the rate of decomposition of H2O2. [1] OR Since O 2 evolved when NaOH(aq) is added, H 2O2 is being oxidised in an alkaline medium. This implies that H 2O2 is a stronger reducing agent in an alkaline medium. OR H2O2 decomposes more rapidly in alkaline medium as compared to in acidic medium. 2 (a) Headers and units for transfer time and t Transfer time in min & s, to nearest s. t correctly calculated to 1 d.p. [1] Headers and units for burette readings and volume of FA 5 Correct calculation of volume of FA5 added All burette readings and volumes of FA5 to 2 d.p. (0.05 cm3). [1] 5 sets of data, 1st and 2nd aliquot within 1st and 2nd min of expt (accept up to 1.9 and 2.9 min), last aliquot not exceeding 10 min [1]

2024 HCI C2 H2 Chemistry Prelims / Paper 4 Example Transfer time t /min Final burette reading /cm3 Initial burette reading /cm3 Volume added /cm3 min s 1 01 1.0 16.50 0.00 16.50 2 00 2.0 27.25 16.50 10.75 4 00 4.0 32.75 27.25 5.50 6 00 6.0 35.90 32.75 3.15 8 00 8.0 37.85 35.90 1.95 2 (b) Label axes & units, suitable scale (not odd, plots occupy at least half the grid in x and y directions). [1] All points correctly plotted within ½ small square. [1] Suitable best-fit curve ignoring anomalous points. [1] Example t½ values: 1.7 min (16.0 cm3 to 8.0 cm3); 1.9 min (12.0 cm3 to 6.0 cm3); 2.4 min (8.0 cm3 to 4.0 cm3) 0.00 4.00 8.00 12.00 16.00 20.00 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Volume of FA 5 /cm3 t /min

2 2 (c) Show working for at least two half-lives. [1] Conclude the order of reaction based on whether t½ is (approximately) constant. [1] E.g. Since volume of FA 5 (VFA5) is directly proportional to concentration of MnO4− in the reaction mixture, for V FA5 to decrease from 16.0 cm3 to 8.0 cm3, the half-life is 1.7 min. For VFA5 to decrease from 12.0 cm3 to 6.0 cm3, the half-life is 1.9 min. Since the two half-lives are almost identical, I agree that the order with respect to MnO4− is one. 2 (d) (i) Draws tangent at t = 5.0 min, tangent line must touch the curve at t = 5.0 min [1] An example is drawn on the graph in 2(b). (ii) Gradient of tangent = (7.6 – 0.0) / (2.0 – 8.6) = –1.15 cm3 min–1 (answer must be negative) [1] (iii) rate of change in amount of S2O32– ions = –1.152  10–3  0.0050 = –5.76  10–6 mol min–1 [1] 2 (d) (iv) rate of change in amount of MnO4– ions in 10 cm3 = –5.76  10–6  1/2  2/5 = –1.15  10–6 mol min–1 [1] (v) rate of change of the concentration of MnO4– ions = (–1.15  10–6) / 0.010 = –1.15  10–4 mol dm–3 min–1 [1] 2 (e) (i) Sigmoidal, showing the gradient becoming more negative (or increase in magnitude) before the point of inflexion and then becoming less negative (or decrease in magnitude) after that. [1] Reaction starts slowly at the beginning, then speeds up, then slows down at the end. [1] (ii) Without adding FA 1, the reaction is slow initially (as both MnO4– and C2O42– are negative ions. This leads to a high activation barrier ); the Mn2+ produced (auto)catalyses the reaction explaining the increase in reaction rate/ increase in magnitude of gradient of the curve after some time. [1] For the graph in 2(b), FA 1 is added in step 3. FA 1 contains the catalyst Mn2+ and the reaction proceeds via the alternative catalysed pathway, causing reaction rate to be high right from the start of the reaction. [1] 3 (a) (i) Correct calculations of each titre value. Mark lost if initial and final readings are swapped. [1]

2024 HCI C2 H2 Chemistry Prelims / Paper 4 Correct recording of titration values to nearest 0.05 cm3 [1] At least 2 titration results consistent to within 0.10 cm3 [1] (ii) Correctly picks 2 values for calculation of average titre value based on the hierarchy: - 2 identical values - average of titre values within 0.05 cm3 - average of titre values within 0.10 cm3 Correct calculation for average titre value to 2 d.p. [1] Accuracy Teachers’ value: 18.80 cm3 18.70 – 18.90 cm3 = [2] 18.60 – 19.00 cm3 = [1] 3 (b) (i) Na2CO3 + H2SO4 → Na2SO4 + CO2 + H2O [1] (ii) Calculation of amount of carbonate in 25.0 cm3 [1] Calculation of percentage purity by mass [1] Example: Average titre value = 19.00 cm3 Amount of acid use = 19.00/1000 × 1.00 = 0.0190 mol Amount of carbonate in 25.0 cm3 = 0.0190 mol ecf from balanced equation in 1(b)(i) Amount of carbonate in 1 dm3 = 0.0190 / 25 x 1000 = 0.760 mol Mass of sodium carbonate = 0.760 × 106 = 80.56 g Percentage purity by mass = 80.56/90 × 100 = 89.51 % (ecf from wrong value for amount of carbonate in 25.0 cm3) Correct s ignificant figures for all final answers in calculation questions for whole paper Correct units in final answer for Q3. [1] Show working for all calculation questions attempted for the whole paper. No penalty for questions left blank. [1] 4 (a) Heat lost by hot water = mc∆T = 75 x 4.18 x (60.4 – 25.6) = 10909.8 J [1] Heat gained by cold water = mc∆T = 75 x 4.18 x (25.6 – 8.9) = 5235.45 J [1] Difference in value of heat = 10909.8 – 5235.45 = 5674 J Heat capacity of copper can = (5674)  (25.6 – 8.9) = 340 J K−1 [1]

3 4 (b) Labelled diagram: Sample procedure: 1. Weigh the spirit burner with ethanol using an electronic balance. Record its mass, M1. 2. Using a measuring cylinder, measure 150 cm3 of water and pour into the copper can. 3. Record the initial temperature of the water, T1, with a thermometer. 4. Set up the apparatus as shown in the diagram