2023 RI H2 Chem Prelims P4 Answers

2023 H2 Chemistry Preliminary Exams Paper 4 – Suggested Solutions Qn 1 (a) Expt Volume of FA 1 / cm3 Volume of deionised water / cm3 t / s part (c) rate / mol dm−3 s−1 1 5.0 25.0 29.0 2.46 x 10−5 2 8.0 22.0 16.2 4.41 x 10−5 3 11.0 19.0 12.2 5.85 x 10−5 4 15.0 15.0 9.4 7.60 x 10−5 (b) Amount of S2O32− added = 0.00500 × 20 1000 = 1.000 × 10-4 mol Amount of I2 reacted with S2O32− added = 1 2 x 1.000 × 10-4 = 5.00 × 10-5 mol = Amount of I2 produced just before the first appearance of blue-black colour Concentration of I2 in reaction mixture = 5.00 × 10-5 ÷ 70 1000 = 7.14 × 10-4 mol dm−3 (c) See table in part (a).

(d)(i) (d)(ii) Since total volume of reaction mixture is kept constant, volume of FA 1 is directly proportional to [Fe 3+] in the mixture. Since the graph is a straight li ne with a pos itive gradient, rate is directly p roportional to [Fe 3+], the reaction is first order with respect to Fe3+. (e) Agree with student. The reaction time would be longer since more time is required to produce more iodine to react with the larger amount of thiosulfate ions present , and so the percentage error is reduced. (f)(i) rate = k[Fe3+][I−]2 (f)(ii) From the slow step, the rate equation is: rate = k’[[FeI]2+][I−] = k[Fe3+][I−]2. This matches the rate equation for the reaction. Hence the proposed mechanism is consistent with the observed kinetics data. 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 Volume of FA 1/ cm3 rate / 10−5 mol dm−3 s−1

2 Qn 2 (a)(i) Table 2.1 test observations 1 Test solution FA 4 with Universal Indicator paper. UI paper turned orange, pH is 3 2 To 1 cm depth of FA 4 in a clean test- tube, add 1 cm depth of aqueous sodium carbonate. White ppt formed. Effervescence observed. CO2 gas evolved gave a white ppt in limewater. 3 Add 1 cm depth of FA 4 into a clean test-tube. Add aqueous sodium hydroxide, slowly with shaking, until no further change is seen. White ppt. formed, soluble in excess NaOH(aq) to give a colourless solution. (a)(ii) Cation: Al3+ Evidence: In test 2, FA 4 reacted with Na 2CO3 to form a white ppt of Al(OH)3 together with the effervescence of CO2 gas. OR From test 1 / test 2, FA 4 contained an acidic ion. Additionally, in test 3, FA 4 reacted with NaOH to form a white ppt of Al(OH)3 which was soluble in excess NaOH to form the colourless complex ion, [Al(OH)4]−. test observations (a)(iii) To 1 cm depth of FA 4, add a few drops of AgNO3. To 1 cm depth of FA 4 in a boiling tube, add 1 cm depth of aqueous NaOH and a piece of Al foil. Heat cautiously. No ppt / No observable change White ppt formed, soluble in excess NaOH to form a colourless solution. Effervescence of H2 was observed. Upon heating, pungent NH3 evolved turned damp red litmus paper blue. (a)(iv) Anion: NO3− (b)(i) Table 2.2 test observations 1 To 1 cm depth of neutral iron( III) chloride, add 10 drops of FA 5 and shake the test-tube. This solution is FA 6. Proceed to test 2 using this solution. Solution turned violet.

2 To the solution of FA 6, add aqueous sodium hydroxide dropwise, with shaking, until no further change is seen. Violet solution decolourised. Red-brown ppt formed, insoluble in excess NaOH(aq). (b)(ii) Phenol (b)(iii) (b)(iv) [Fe(H2O)6]3+ + L2− ⇌ [Fe(H2O)4L]+ + 2H2O (b)(v) When NaOH was added, the red -brown ppt of Fe(OH) 3 was formed. Thus, the concentration of [Fe(H2O)6]3+ decreases and the position of equilibrium of the equation in (iv) shifted to the left, decreasing the concentration of [Fe(H2O)4L]+ and causing the violet solution to decolourise. Qn 3 (a)(i) Dilution of FA 7 Final burette reading / cm3 30.50 Initial burette reading / cm3 0.00 Volume of FA 7 used / cm3 30.50 Titration results Titration number 1 2 Final burette reading / cm3 24.20 24.20 Initial burette reading / cm3 0.00 0.00 Volume of FA 8 used / cm3 24.20 24.20 Values used (✓) ✓ ✓ (a)(ii) Average volume of FA 8 used = 24.20 + 24.20 2 = 24.20 cm3 (2 d.p.) (b)(i) Amount of MnO4– ions = 24.20 1000 x 0.00500 = 1.21 x 10–4 mol (3 s.f.) (b)(ii) Amount of Fe2+ ions in 250 cm3 of FA 10 = 1.21 x 10–4 x 5 x 250 25.0 = 6.05 x 10–3 mol (3 s.f.) (b)(iii) Amount of Fe2+ ions in 250 cm3 of FA 7 = 6.05 x 10–3 x 250 30.50 = 0.04959 mol = 0.0496 mol (3 s.f.) (b)(iv) Either Amount of FeO in magnetite = 0.04959 mol Mass of FeO = 0.04959 x (55.8 + 16.0) = 3.5606 g Amount of Fe2O3 in magnetite = 0.04959 mol Or Amount of magnetite, Fe3O4 = 0.04959 mol Mass of magnetite in sample A = 0.04959 x (3(55.8) + 4(16.0)) = 11.4752 g

4 Mass of Fe2O3 = 0.04959 x (2(55.8) + 3(16.0)) = 7.9146 g Mass of magnetite in sample A = 3.5606 + 7.9146 = 11.4752 g Percentage by mass of magnetite in sample A = 11.4752 15.0 x 100 = 76.5% (3 s.f.) (c)(i) Ecell = +1.33 – (0.77) = +0.56 V Since Ecell is > 0, reaction is feasible. (c)(ii) The titration result may be inaccurately determined as the end-point colour change is not distinct. It is difficult to detect one drop of excess orange K2Cr2O7 solution in a yellow-green solution (containing Fe3+ & Cr3+). (d)(i) Percentage error = 6.03-5.95 6.03 x 100 = 1.33% (d)(ii) The student performed the experiment well as the calculated percentage error is less than the maximum percentage error of the whole experimental procedure. Qn 4 (a) Let V cm3 be the volume of solution B (containing Al3+) and (60.0 − V) cm3 be the volume of solution Z (containing Na+) used at which stoichiometric amount of Al3+ and Na+ react. Amount of Al3+ in V cm3 of solution B = Amount of Na+ in (60.0 − V) cm3 of solution Z V 1000 x 0.120 = 60 - V 1000 x 0.075 0.120V = 4.5 - 0.075V V = 23.1 cm3 Approximate equivalence volume of solution B = 23.1 cm3 (b) Procedure Experiment volume of solutions added / cm3 initial temperature of solution Z / ºC highest temperature of resultant solution / oC T / oC Z B 1 50.0 10.0 2 45.0 15.0 3 42.0 18.0 4 40.0 20.0 5 38.0 22.0 6 36.0 24.0 7 34.0 26.0 8 30.0 30.0 9 25.0 35.0 10 20.0 40.0

1. Using a 50 cm3 measuring cylinder, measure 50.0 cm3 of solution Z into a clean and dry Styrofoam cup supported in a 250 cm3 beaker. 2. Use a thermometer to measure and record the steady initial temperature of solution Z. 3. Using another 50 cm3 measuring cylinder, measure 10.0 cm3 of solution B. 4. Add solution B from the measuring cylinder to solution Z in the cup, u se the thermometer to stir the mixture gently. Measure and record the highest temperature of the resultant solution. 5. Empty, wash and carefully dry the cup. 6. Repeat steps 1 to 5 using volumes of solution Z and solution B in the table above so that the total volume of the mixture is 60.0 cm3. 7. Record the temperature increase, T, for each experiment by taking highest temperature – initial temperature. (c) T / oC 0 volume of solution B / cm3 (d) Stoichiometric amount of Al3+ to react with Na+ = Veq 1000 x 0.120 mol