HCI 2026 H2 Chemistry Prelim P4 Mark Scheme

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 1 HWA CHONG INSTITUTION 2026 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION PAPER 4 MARK SCHEME 1 (a) 1 2 3 4 5 6 Obs 1. red-brown ppt formed Obs 2. insoluble in excess Obs 3. red-brown residue Obs 4. colourless filtrate Obs 5. no gas evolved (on warming) Obs 6. white ppt formed insoluble in excess (on adding H2SO4) Obs 7. (red-brown residue dissolves to give) yellow solution Obs 8. solution turns darker yellow/orange/brown ( on adding KI) Obs 9. solution decolourises (on addition of thiosulfate) Obs 10. solution turns purple/violet/brown/grey Obs 11. solution turns colourless (on standing) Obs 12. green ppt (turns brown on standing) (on adding NaOH) Obs 13. insoluble in excess Obs 14. green ppt turns reddish brown (on adding H2O2) Obs 15. effervescence of gas that relights a glowing splint Obs 16. gas is O2 15 – 16 = [8] 13 – 14 = [7] 11 – 12 = [6] 9 – 10 = [5] 7 – 8 = [4] 5 – 6 = [3] 3 – 4 = [2] 1 – 2 = [1] 0 = [0] M1 M2 M3 M4 M5 M6 M7 M8 1 (b) (i) Ba2+, Fe3+ [1] for each M9 M10

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 2 1 (b) (ii) Role of KI: It is a reducing agent. Solution turned brown, showing that I– has been oxidised to I2, this means that Fe3+ has been reduced. [1] Role of H2O2: Oxidising agent. When added to green ppt of Fe(OH) 2, it oxidised Fe(OH)2 to Fe(OH)3 which is the red-brown ppt. [1] M11 M12 1 (c) Obs 1. white ppt formed Obs 2. ppt dissolves in ammonia (to give a colourless solution) [1] for each observation M13 M14 1 (d) Cl– [1] M15

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 3 2 (a) (b) (i) All the burette readings are recorded to the nearest 0.05 cm3. ➢ volume of FA 3 calculated correctly & final and initial burette readings are not inverted ➢ Volume of FA 2 used is between 49.00 - 49.50 cm3 M16 At least two consistent titre readings of end-point within 0.10 cm3 M17 Sample results: (a) Final burette reading / cm3 49.50 Initial burette reading / cm3 0.00 Volume of FA 2 used / cm3 49.50 (b)(i) 1 2 Final burette reading / cm3 20.90 20.90 Initial burette reading / cm3 0.00 0.00 Volume of FA 5 used / cm3 20.90 20.90 ✔ ✔ Calculate the Teacher’s and the student’s scaled mean titre values using the following expression: Scaled mean titre = 49.25  mean titre volume of FA 2 diluted Calculate the difference between the student’s scaled mean titre value and the Teacher’s scaled mean titre value. Teacher’s scaled mean titre = 20.79 cm3 Give 2 marks if this difference is ≤ 0.3 Give 1 mark if this difference is > 0.3 but ≤ 0.5 Give 0 marks if this difference is > 0.5 M18 M19 2 (b) (ii) working shown using two consistent titres & average calculated to 2 d.p. with units stated ➢ Mark is lost if the titres used are not identified either in the table (by, for example, a tick) or in a calculation or identified wrongly or no titre within 0.10 cm3. ➢ Mark is lost if there are arithmetic errors in the titration results table. ➢ Mark is lost if the candidate made an arithmetic error in the calculation of the mean titre. M20

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 4 2 (d) (i) nO2 produced by 1cm3 of H2O2 = 0.75 ÷ 24000 = 3.125 × 10−5 mol nH2O2 in 1 cm3 = 3.125 × 10−5 × 2 = 6.25 × 10−5 mol [H2O2] = 6.25 × 10−5 ÷ 1/1000 = 0.0625 mol dm−3 M21 M22 2 (d) (ii) nH2O2 in 25.0 cm3 = 0.0625 × 25/1000 = 1.56 × 10−3 mol = nI2 nS2O32− = 1.56 × 10−3 × 2 = 3.125 × 10−3 mol [S2O32−] = 3.125 × 10−3 ÷ 31.50/1000 = 0.0992 mol dm−3 M23 M24 2 (d) (iii) 1. nS2O32− = 0.0992(or ecf) × (titre in (b)(ii) ÷ 1000) 2. nKIO3 = nS2O32− ÷ 6 3. [KIO3] in FA 6 = nKIO3 ÷ 25/1000 Award 1 mark for the correct use of the mole ratio in step 2 and 1 mark for steps 1 and 3 if both correct (allow ecf for step 3 if step 2 is wrong). M25 M26 2 (d) (iv) [KIO3] in FA 2 = ANS(d(iii)) × (250 ÷ volume of FA2 used) M27 Correct s.f. (3 s.f) for final answer from (d)(i) to (iv) M28

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 5 3 (a) Correct headers and units M29 Correct precision • Time to nearest second • 1/t to 3 s.f • Volume reading to 1 d.p. (FA3, FA7, FA8, Deionised water) M30 Correctly calculates 1/t M31 Volumes of FA7 chosen are well spaced over 5 - 25 cm3 M32 • Total volume kept constant by adding deionised water • Constant volume of FA 3 and FA 8 M33 Sample results: Experiment No. Volume / cm3 Time / s time 1 /s–1 FA 8 FA 7 FA 3 Deionised water 1 50.0 25.0 25.0 0.0 10 0.100 3 50.0 20.0 25.0 5.0 13 0.0769 4 50.0 15.0 25.0 10.0 17 0.0588 5 50.0 10.0 25.0 15.0 26 0.0385 2 50.0 5.0 25.0 20.0 54 0.0185 3 (b) If total volume is kept constant, vol (H 2O2) used would be directly proportional to [H2O2] in the reaction mixture. M34 3 (c) • Labels both axes with headings and units and uses appropriate scale • All points plotted to within half small square • Obtains best-fit straight line passing through the origin / almost passing through origin M35 M36 M37

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 6 Sample graph: 3 (d) Graph of 1/time vs vol (H2O2) is a straight line passing through the origin/ almost passing through zero  rate  [H2O2]  first order wrt [H2O2] M38 M39 3 (e) rate = k [H2O2][I−][H+] units of k = mol dm-3 s-1 (mol dm-3) 3 = mol–2 dm6 s–1 ecf from (d) M40

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 7 4 (a) [Ba(OH)2]  Volume Ba(OH)2 = 1 [CH3CO2H]  Volume CH3CO2H 2 Given [CH3CO2H]  [Ba(OH)2]  Volume Ba(OH)2 : Volume CH3CO2H = 1:2 Volume of Ba(OH)2 = 50 × 1/3 = 17 cm3 Volume of CH3CO2H = 50 × 2/3 = 33 cm3 or use algebra: Let volume Ba(OH)2 = V cm3, then volume CH3CO2H = (50 – V) cm3 [Ba(OH)2]  V = 1 [CH3CO2H]  (50 – V) 2 Volume of Ba(OH)2 = V = 17 cm3 Volume of CH3CO2H = 50 – V = 33 cm3 [0.5] × 2 for each volume, including working M41 4 (b) M42: A table showing all the chosen different volumes of the two solutions to be used. Total volume must be 50 cm3. There should be at least three Ba(OH)2 volumes < 17 cm3 and at least three Ba(OH)2 volumes > 17 cm3 so that there are at least three plotted points on each line. M43: Same table showing measurement of initial temperature of Ba(OH)2 (or CH3CO2H), the highest temperature reached after mixing and T. M44: Use of thermometer, two measuring cylinders. Burette can be used for the first reagent added, but second reagent added must be measuring cylinder. However, student is not penalised here for apparatus if they used burettes for both reagents. M46 will be lost instead. M45: Took action to minimise heat loss e.g. using two styrofoam cups stacking one into the other / use a lid for the styrofoam cup / switch off the fans to ensure draft -free environment. M45 is lost if student uses a beaker as the reaction container (instead of styrofoam cup / polystyrene cup). Using just one styrofoam cup is not good enough to score M45. M46: Overall sequence of actions: M42 M43 M44 M45 M46

2026 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 8 measure volume of one solution, add quickly into a styrofoam cup  measure temperature of this solution in the cup  measure volume of second solution, add it into the same cup  stir and measure the highest temperature r