RI 2026 H2 Chem Prelims P4 Answers

© Raffles Institution 2026 9476/04/S/26 2026 RI YEAR 6 PRELIMINARY EXAMINATION Paper 4 Practical [Answers] 1 (a)(i) Titration results Final burette reading / cm3 22.00 42.50 Initial burette reading / cm3 0.00 20.50 Volume of FA 2 used / cm3 22.00 22.00 Values used (✓) ✓ ✓ 1 (a)(ii) Average volume of FA 2 used = 22.00 + 22.00 2 = 22.00 cm3 (2d.p.) 1 (b)(i) Amount of NaOH used = 22.00 1000  0.100 = 2.20  10−3 mol 1 (b)(ii) Amount of H+ in HCl and CH3COOH = Amount of OH− used = 2.20  10−3 mol Total [HCl] + [CH3COOH] = 2.20 10-3 25.0 1000 = 0.0880 mol dm−3 1 (b)(iii) [CH3COOH] in FA 1 = 0.0880 – 0.0300 = 0.0580 mol dm−3 1 (b)(iv) K1 = [CH3COOH][ROH] [CH3COOR] 1 (b)(v) CH3COOR(aq) + H2O(l) ⇌ CH3COOH(aq) + ROH(aq) Initial conc / mol 0.0600 − 0 0 Change in conc / mol −0.0580 − +0.0580 +0.0580 Equilibrium conc / mol 0.002 − 0.0580 0.0580 K1 = 0.0580  0.0580 0.002 = 1.68 mol dm−3 1 (c) K2 = [CH3COOR][H2O] [CH3COOH][ROH] [H2O] is not constant since it is a product and not a solvent . Therefore, the student is incorrect.

© Raffles Institution 2026 9476/04/S/26 1 (d) Maximum total % error = ( ±0.06 25.0 + 2(±0.05) 20.00 )  100% = ±0.740% 1 (e)(i) test observations with FA 3 1 Add 1 cm depth of FA 3 into a clean test-tube. To this test-tube add 10 drops of sodium hydroxide followed by iodine solution, dropwise, until a permanent yellow/orange colour is present. Warm the test-tube in the beaker of hot water for 2 minutes. Yellow ppt formed. Orange solution decolourised. 2 Add 1 cm depth of FA 3 into a clean test-tube. Add 8 drops of Fehling’s solution. Warm the test-tube in the beaker of hot water for 3 minutes. Reddish-brown / brick-red ppt observed. 3 Test solution FA 3 with Universal Indicator paper. Universal Indicator paper turns green / yellow-green. pH 6 / 7 1 (e)(ii)

© Raffles Institution 2026 9476/04/S/26 2 (a) tests observations 1 Add 1 cm depth of FA 4 into a clean test-tube. Add 1 cm depth of aqueous sodium carbonate. *White and pink ppt/ light pink/ pink ppt formed in a pink solution. Effervescence of CO2 which gave a white ppt with limewater is produced. 2 Add 1 cm depth of FA 4 into a clean test-tube. Add 2 cm depth of aqueous sodium hydroxide to the mixture. Filter the mixture. Add dilute nitric acid dropwise to the filtrate until no further change. *White and pink ppt/ light pink/ pink ppt formed. White ppt dissolved in excess sodium hydroxide, leaving behind a pink ppt in a colourless solution. **Pink/pale pink/pale purple residue and colourless filtrate. White ppt reformed which dissolved in excess dilute nitric acid to form a colourless solution. 3 Add 1 cm depth of FA 4 into a clean test-tube. Add 1 cm depth of aqueous silver nitrate. Add 2 cm depth of aqueous ammonia to the mixture. To a separate clean test- tube, add 1 cm depth dilute nitric acid. Filter the mixture into this dilute nitric acid. White ppt. White and pink ppt insoluble in excess aqueous ammonia. Pink residue and colourless filtrate. White ppt formed in dilute nitric acid. 2 (b) cation present Al3+ anion present Cl– 2 (c) Ag+(aq) + Cl–(aq) ⇌ AgCl(s) … (1) Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)2]+ (aq)… (2) On filtering into dilute nitric acid, the acid reacts with aqueous ammonia. This causes [NH3] to decrease and position of equilibrium of (2) shifts to the left, increasing [Ag+]. This causes position of equilibrium (1) to shift to the right, reforming AgCl precipitate.

© Raffles Institution 2026 9476/04/S/26 3 (a) VFA7 / cm3 Vwater / cm3 t / s log(100 t ) log(VFA7) 1 8.0 14.0 42 0.377 0.903 2 10.0 12.0 33 0.481 1.00 3 12.0 10.0 27 0.569 1.08 4 16.0 6.0 20 0.699 1.20 3 (b)(i) 3 (b)(ii) Using coordinates (1.20, 0.70) and (1.00, 0.48), Gradient = 0.70 − 0.48 1.20 − 1.00 = 1.10 Order of reaction with respect to S2O82− = 1 3 (c) VFA4 / cm3 Vwater / cm3 VFA7 / cm3 t / s 8.0 6.0 8.0 27 3 (d) 2M2+(aq) + S2O82−(aq) ⎯→ 2M3+(aq) + 2SO42−(aq) M2+ reduces S2O82− to SO42− and is oxidised to M3+(aq) 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25 log (100/t) log (VFA7) (1.20,0.70) (1.00,0.48)

© Raffles Institution 2026 9476/04/S/26 2M3+(aq) + 2I–(aq) ⎯→ 2M2+(aq) + I2(aq) M3+(aq) oxidises I–(aq) to I2 and is reduced to M2+(aq). Each step involves a reaction between oppositely charged ions which attract each other. Hence M2+ acts as a homogenous catalyst by offering an alternative reaction pathway with a lower activation energy. 4 (a) Heat evolved for temp. rise of 5 oC = 100  4.18  5 = 2090 J Amount of CuSO4 required = 2090 65.0×103 = 0.03215 mol Minimum mass of CuSO4 to use = 0.03215 x 159.6 = 5.13 g (3 sf) 4 (b) 1. Using an electronic balance, weigh accurately about 6.00 g of CuSO4 into a clean and dry weighing bottle. 2. Record the total mass of CuSO4 and the weighing bottle, m1. 3. Using a 100 cm3 measuring cylinder, transfer 100 cm3 of water into a dry Styrofoam cup, placed in a 250 cm3 beaker. 4. Place the lid on the cup, slip the thermometer through the lid and stir the water gently using the thermometer. 5. Record the initial temperature of the water. Start the stop -watch and record the temperature of the water every 30 seconds for 2.5 min. 6. At exactly 3 min, add the CuSO4 in the weighing bottle to the Styrofoam cup. Cover the cup with a lid and continue to stir the mixture gently with the thermometer and record the temperature every 30s from 3.5 min to 10.0 min. 7. Reweigh the emptied weighing bottle and record its mass, m2. Take the difference in mass between m1 and m2 and hence record the actual mass of CuSO4 that was transferred and dissolved in water. 8. Plot graph of temperature against time. 9. Extrapolate the graph to the third minute when CuSO4 was added to water to obtain the highest temperature reached. 10. Repeat steps 1 to 9 to obtain another set of data to plot another graph. Obtain a second value of Tmax and enthalpy change of reaction and find the average value for the enthalpy change of reaction.

© Raffles Institution 2026 9476/04/S/26 Hhydration 4 (c)(i) 4 (d) CuSO4(s) + 5H2O(l) CuSO4•5H2O(s) Cu2+(aq) + SO42–(aq) + 5H2O(l) Hhydration = –65.0 – (+12.0 ) = −77.0 kJ mol−1 +12.0 kJ mol−1 −65.0 kJ mol−1