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