HCI Prelim P4 Answers (For sharing)

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 HWA CHONG INSTITUTION 2020 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 4 1 (a) Correct headings and units AND initial and final burette readings recorded [1] All burette readings to 0.05 cm3 AND correct calculation of all titre volumes [1] 1 (b) Consistent titres indicated with ticks or shown in the calculation of the average titre AND correct calculation of average titre [1] Accuracy marks: Award [2] if 24.70 cm3  average titre  25.10 cm3 Award [1] if 24.60 cm3  average titre < 24.70 cm3 Award [1] if 25.10 cm3 < average titre  25.20 cm3 Otherwise, no marks awarded for accuracy 1 (c) (i) Amount of sodium hydroxide = (average titre) ÷ 1000 × 0.0400 = A mol [1] 1 (c) (ii) Amount of ethanedioic acid = A ÷ 2 = B mol [1] 1 (d) 2 MnO4−(aq) + 16 H+(aq) + 5 C2O42−(aq) 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l) [1] 1 (e) (i) Amount of manganate(VII) ions = 24.80 ÷ 1000 × 0.0200 = 4.96 × 10−4 mol [1] 1 (e) (ii) Amount of ethanedioate ions = 4.96 × 10−4 × 5 ÷ 2 = 0.00124 mol [1] 1 (f) Amount of sodium ethanedioate = 0.00124 – B = C mol [1] 1 (g) Mass of ethanedioic acid = B × 90 = D g Mass of sodium ethanedioate = C × 134 = E g [1] Percentage by mass of ethanedioic acid = D ÷ (D+E) × 100 % [1] 2 (a) Mass of capped bottle and solid /g 6.65 Mass of empty bottle and cap /g 5.39 Mass of solid used /g 1.26 [1] headers and units and all readings to 2 d.p.

1 2 (b) FA 3 FA 4 NaOH No ppt On warming (or heating), gas evolved (or gas produced) turned damp red litmus paper blue White ppt insoluble in excess NH3 No ppt White ppt insoluble in excess 5 correct observations  3m 3-4 correct observations  2m 1-2 correct observations  1m 2 (c) reagents used: AgNO3(aq) followed by NH3(aq) unknown observations FA 3 cream/pale cream ppt partially soluble in NH3 Accept: cream/pale cream ppt insoluble in NH3 Accept: pale yellow ppt partially soluble in NH3 FA 4 no ppt [1] for reagents used [1] for both sets of observations 2 (d) unknown observation FA 5 purple KMnO4 decolourised or turned colourless or purple solution decolourised or turned colourless [1] 2 (e) [1] for each ion in the correct FA and explanation with reference to the observation in 2(b) and ppt formula stated i.e. AgBr, and Mg(OH)2 FA 3 contains NH4+ and Br– Explanation: NH4+ gives NH3 gas on warming/heating with NaOH(aq). Br– gives cream AgBr ppt with Ag +(aq) and ppt partially soluble ( accept: insoluble) in NH3(aq). FA 4 contains Mg2+ Explanation: White Mg(OH) 2 ppt produced with NaOH(aq) and with NH3(aq), and ppt insoluble in excess reagent. FA 5 contains SO32–

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 Explanation: FA 5 contains a sulfur-containing anion (pg19) which would be SO42– or SO 32–. SO 32– can be oxidised (to SO42–) so SO32– can reduce and therefore decolourise KMnO4 (in acid solution). 2 (f) [1] for observation: White ppt dissolves [1] for explanation and balanced equation: The added NH 4+, from NH 4Cl, reacts with the Mg(OH)2 ppt, which hence dissolves. Mg(OH)2 + 2NH4+ → Mg2+ + 2H2O + 2NH3 or: Mixture P contains the equilibrium, Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH–(aq) The added NH4+ removes OH–, NH4+ + OH– → NH3 + H2O The position of the above equilibrium shifts to the right, therefore the white ppt Mg(OH)2(s) dissolves (or in terms of [OH–] decreases, ionic product for Mg(OH)2 becomes less than its Ksp) 2 (g) [1] for each ion – the chosen reagent(s) and how to decide if the test result is positive (i.e. expected observation & conclusion of anion) test 1: Add H2SO4(aq) / HCl(aq) / HNO3(aq) If effervescence is seen and the gas gives white ppt with limewater, the ion is CO32–. If a (pale) brown gas is seen, the ion is NO2–. test 2: Add NaOH(aq) and Al foil and heat If a gas is produced (NH3 gas) that turns damp red litmus paper blue, the ion is NO3–. NO2– also gives NH 3 gas in test 2 but it would already have giv en a positive result in test 1. or test 1: Add NaOH(aq) and Al foil and heat If a gas is produced (NH3 gas) that turns damp red litmus paper blue, the ion is NO3– or NO2–. test 2: Add H2SO4(aq) / HCl(aq) / HNO3(aq) If effervescence is seen and the gas gives white ppt with limewater, the ion is CO32–. If a (pale) brown gas is seen, the ion is NO2–. So if the FA solution gives positive test in test 1 but negative test in test 2, the ion is NO3–.

2 3 (a) (i) (ii) expt volume of FA 5 / cm3 volume of FA 6 / cm3 volume of water in solution 1 / cm3 t / s 1/t / s–1 lg(1/t) lg(VFA 6) 1 15.0 85.0 0.0 32 2 15.0 25.0 60.0 10 3 4 5 [1] Table with correct headers and units. [1] Record all data with correct precision  volumes to 0.5 cm3  t to nearest second  calculated values to 3 sf. [1] Complete set of volume of FA 6 and time readings for 5 experiments and all values of t increase as volume of FA 6 decreases. Data for experiments 1 and 2 must be included. [1] Choose 3 other well-spaced values for the volume of FA 6. All volumes must differ by at least 10 cm3. [1] Correctly calculates 1/t, lg(1/t) and lg(VFA 6) values for all 5 experiments. 3 (b) (i) [1] Axes correct way round, with correct labels and appropriate scale. Scale must be chosen so that plotted points occupy at least half the graph grid in both x and y directions. [1] All plotted points correct to within ± ½ small square. [1] A best fit line is drawn with anomalous points excluded. 3 (b) (ii) [1] Correct calculation of gradient of the graph with working shown. The m value is equal to the gradient of the line. 3 (c)  Correct calculation of lg(1/25) = –1.40  Correct reading of lg(VFA 6) value from graph to within ±½ small square.  Correct calculation of volume of FA 6 using value for lg(VFA 6). [2] for all 3 points correct. [1] for 2 points correct. 3 (d) Selects Experiment 1 and explains that time, t, is smallest value and so has greatest % error. [1] OR Selects Experiment 2 and explains that volume of FA 6 used (25 cm 3) is smallest and so has greatest % error [1]

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 3 (e) Calculates n(H+) = 0.1 × 2 × 85 × 10–3 = 0.0170 mol [H+] = 0.0170 / (205/1000) = 0.0829 mol dm–3 [1] Shows correct units in final answer in part 3(c) and appropriate significant figures in all final answers: 3 s.f. in 3(b)(ii), 3 or 4 sf in 1(c), 1(e), 1(f), 1(g), 3(c), 3(e) and 2 d.p. in 1(b) and attempting 1(b), 1(c), 1(e), 1(f), 1(g), 3(b)(ii), 3(c), 3(e) 3 (f) (i) The value of t will be lower. This is because the [H+] is higher than expected, so rate of reaction is faster. 3 (f) (ii) Measuring volumes of FA 5, FA 6 and water separately and mixing to make solution 1. 4 (a) [H2C2O4]  Volume H2C2O4 = 1 [NaOH]  Volume NaOH 2 Given [H2C2O4]  [NaOH]  Volume H2C2O4 : Volume NaOH = 1:2 Volume of H2C2O4 = 50 × 1/3 = 17 cm3 Volume of NaOH = 50 × 2/3 = 33 cm3 or use algebra: Let volume H2C2O4 = V cm3, then volume NaOH = (50 – V) cm3 [H2C2O4]  V = 1 [NaOH]  (50 – V) 2 Volume of H2C2O4 = V = 17 cm3 Volume of NaOH = 50 – V = 33 cm3 [1] for both volumes, including working 4 (b) M1: A table showing all the chosen different volumes of the two solutions to be used. Total volume must be 50 cm 3. There should be at least three H2C2O4 volumes < 17 cm 3 and at least three > 17 cm 3 so that there are at least three plotted points on each line.  M1 and M5 are lost if the student adds extra water / uses same acid volumes throughout / same NaOH volumes throughout.  M1 is lost if there is no tabulation of all the chosen volumes – the question wanted this tabulation (table need not have lines).

3  M1 is lost if there is only one or two data points to d