2023 Paper 4Ans

Upon adding KI (aq), yellow solution turns orange Upon standing, orange solution turns brown. On adding FA2, brown solution turns yellow, then green and eventually blue. When FA3 is added, yellow solutions turns green, then blue, then darker blue. Effervescence is observed. (Not required to test for gas) On adding all of FA4, blue solution turns violet/purple. Effervescence is observed. (Not required to test for gas) Red-brown solid is formed. Upon standing, purple solution turns blue. Scan QR code for QA Video 2023 A level H2 Chem Paper 4

2VO2+ + 4H+ + 2I− ⎯→ 2VO2+ + 2H2O + I2 In presence of excess zinc, VO2+ is reduced to V2+. VO2+ is reduced by Zn to VO2+ (Ecell = +1.76 V), VO2+ is reduced by Zn to V3+ (Ecell = +1.10 V), V3+ is reduced by Zn to V2+ (Ecell = +0.50 V). V2+ cannot be reduced by Zn to V (Ecell = −0.44 V) Violet Yellow

Experiment 1 Time / min (1.d.p.) Burette reading / cm3 (1.d.p.) Volume of gas collected / cm3 (1.d.p.) 0.0 49.8 0.0 1.0 39.4 10.4 2.0 28.4 21.4 3.0 16.6 33.2 4.0 8.9 40.9 5.0 7.3 42.5 6.0 7.2 42.6 Experiment 2 Time / min (1.d.p.) Burette reading / cm3 (1.d.p.) Volume of gas collected / cm3 (1.d.p.) 0.0 50.0 0.0 1.0 43.0 7.0 2.0 35.4 14.6 3.0 27.8 22.2 4.0 24.0 26.0 5.0 23.6 26.4 6.0 23.2 26.8

Gradient of tangent (at t = 0) = 33.0 - 0 3.0 - 0 = 11.0 cm3 min−1 Gradient of tangent (at t = 0) = 22.0 - 0 3.0 - 0 = 7.33 cm3 min−1 For expt 1, [HCl]= 0.0800 mol dm−3, for expt 2, [HCl] = 0.08 × 30 50 = 0.480 mol dm−3 When [HCl] × 0.6, rate of reaction × 0.67, hence it is first order w.r.t [HCl].

Titration 1 2 Final burette reading / cm3 15.10 15.10 Initial burette reading / cm3 0.00 0.10 Volume of FA6 used /cm3 15.10 15.00 Volume of FA6 = 15.10 + 15.00 2 = 15.05 cm3 Amount of NaOH added = 1.00 × 25.0 1000 = 0.0250 mol Amount of HCl used = 0.0800 × 15.05 1000 = 0.001204 mol Amount of NaOH in 25 cm3 (withdrawn sample) = 0.001204 mol Amount of NaOH in 250 cm3 (volumetric flask) = 0.001204 × 250 25 = 0.01204 mol Amount of NaOH reacted with aspirin = Total NaOH added − NaOH remaining = 0.0250 − 0.01204 = 0.01296 mol

Total amount of aspirin reacted with NaOH = ½ × 0.01296 = 0.00648 mol (5 tablets) Amount of aspirin in 1 tablet = 0.00648 × 1 5 = 0.001296 mol Mass of aspirin in 1 tablet = 0.001296 × 180 = 0.233 g 0.233 g = 233 mg Percentage error = 233 − 300 300 × 100% = −22.3 %