RVHS 2022 H2 Chemistry Prelim P4 Ans

River Valley High School Pg 18 of 20 JC 2 H2 Chemistry 9729 2022 Preliminary Examination Paper 4 2022 H2 Chemistry Preliminary Examination Paper 4 Suggested Answers 1 Investigation of the oxidising ability of substances (a) (i) Table 1.1 Test Observations 1 To a 1 cm depth of FA 1 in a test-tube, add 1 cm depth of FA 4 , then gradually add FA 2 till 1 drop in excess. Purple FA 2 decolourised. Pale green/colourless FA 1 turned pale yellow and finally orange/ pink. 2 To a 1 cm depth of FA 3 in a test-tube, add 1 cm depth of FA 2. Purple KMnO4 decolourised. A brown solution/ppt formed. Effervescence/ bubbles of gas evolved rapidly. Colourless, odourless gas evolved relight a glowing splint. The gas is oxygen. 3 To a 1 cm depth of FA 1 in a test-tube, add about 1 cm depth of FA 4, followed by 1 cm depth of FA 3. To a portion of resulting solution, add aqueous sodium hydroxide till excess. Pale green /colourless FA 1 turned yellow. Red-brown ppt formed is insoluble in excess NaOH(aq). Effervescence/ bubbles of gas evolved (rapidly). (ii) Purple MnO4− oxidised (pale green) iron(II) to (yellow) iron(III) ions. Itself is reduced to colourless Mn2+. (reject pale Mn2+) (iii) Fe2+ (aq) → Fe3+(aq) + e− Fe3+(aq) + 3OH−(aq) → Fe(OH)3(s) (iv) Compound A / FA 3 (b) (i) Titration results Titration number 1 2 Final burette reading /cm3 24.20 24.25 Initial burette reading /cm3 0.00 1.00

River Valley High School Pg 19 of 20 JC 2 H2 Chemistry 9729 2022 Preliminary Examination Paper 4 Volume of FA 2 (added) /cm3 24.20 24.25 (ii) average volume of FA 2 used = 24.20 24.25 2 + = 24.23 cm3 (c) (i) [KMnO4] = 0.750 1000 39.1 54.9 (4 16.0) 250+ +  = 0.01899 mol dm−3 amount of MnO4− = FA2V0.01899 1000 mol = 0.000460 mol (ii) amount of Fe2+ in 25.0 cm3 = 5(c)(i) 1 mol [Fe2+] = 5 1000(c)(i) 1 25.0 mol dm−3 = 0.0920 mol dm−3 (d) Identify the cause: Chloride is oxidised by/ reacts with MnO4− And any one of the following modification : M16 • The titration needs to be carried out in the fumehood. Chloride is oxidised by MnO4− to give toxic chlorine gas. • Iron(II) chloride needs to be diluted prior to titration. The titre will exceed 50.00 cm3 if iron(II) chloride is not diluted. • Prepare higher concentration of MnO 4− for used. The titre will exceed 50.00 cm3 if FA 2 with original concentration is used. • Using a smaller pipette/ burette, measure a smaller volume of iron( II) chloride for titration, so that the titre will not exceed 50.00 cm3. (5FeCl2(aq) + 3MnO 4−(aq) + 24H +(aq) → 5Fe3+(aq) + 5C l2(g) + 3Mn 2+(aq) + 12H2O(l)) (e) (i) As V FA3 increases, more compound A/FA 3 was added to oxidise Fe 2+ in FA 1. This leaves less Fe2+ to be oxidised by MnO4− in FA 2. (ii) • Not an anomaly. • (Compound A in FA 3 is both an oxidising and reducing agent.) • In experiment 5, compound A is in excess/ Fe2+ is limiting. • The (excess) compound A is oxidised by MnO4−.

River Valley High School Pg 20 of 20 JC 2 H2 Chemistry 9729 2022 Preliminary Examination Paper 4 (iii) percentage uncertainty = 2 0.05 1001.25   = 8.0 % 2 Results Expt VFA 5/ cm3 𝑉H2O /cm3 Reaction time, t /s lg(VFA 5) lg (rate) 1 20.00 0.00 15.1 1.30 2.38 2 10.00 10.00 56.5 1.00 1.80 3 15.00 5.00 26.9 1.18 2.13 4 6.00 14.00 144.5 0.778 1.40 (b) | | | | | | | 0.70 0.80 0.90 1.00 1.10 1.20 1.30 lg(VFA5) lg(rate) 2.40  2.20  2.00  1.80  1.60  1.40  1.20  (0.78, 1.40) (1.21, 2.20) 1.62

River Valley High School Pg 21 of 20 JC 2 H2 Chemistry 9729 2022 Preliminary Examination Paper 4 (c) (i) Gradient of line = 1.40 2.20 0.78 1.21 − − = 1.86 (3 s.f.) m = 2 (nearest integer) (ii) From graph, when lg(7.9) = 0.898, lg( 3600 reaction time ) = 1.62 reaction time = 1.62 3600 10 = 86.4 s (d) (i) Experiment 1: [KI] = 20.000.0500 60.00 = 0.016667 mol dm–3 Experiment 2: [KI] = 10.000.0500 60.00 =0.008333 mol dm–3 20.00 0.016667 210.00 0.008333== (ii) Using results of experiment 1 and 2, 1 2 3600 trate in experiment 1 56.5= 3.74 43600rate in experiment 2 15.1 t = =  When [KI] doubled, rate of experiment 2 is 4 times that of experiment 1. (e) (i) S2O32‒(aq) + 2H+(aq) → S(s) + SO2(aq/g) + H2O(l) (ii) Not as good as that in (a). With any of the following reasons: • There is less thiosulfate left in the reaction mixture to react with iodine formed/ so shorter time recorded. (or words to the effect) • More time is needed to transfer the content in measuring cylinder/ 20.00 cm3 of solution into the beaker for mixing Or As good as that in (a). With any of the following reasons: • [S2O32‒] is very small, so reaction with iron(III) ions/H+/acid will be very slow and negligible. • [S2O32‒] decrease by similar extent for each experiment, the relative rate of experiments is not affected.

River Valley High School Pg 22 of 20 JC 2 H2 Chemistry 9729 2022 Preliminary Examination Paper 4 3 (a) Table 3.1 Test Observations 1 Place a test-tube containing 2 cm depth of FA 9 in an almost boiling water bath for a few minutes. Blue/ bluish -green FA 9 turned green. 2 To a 2 cm depth of FA 9 in a test-tube, add gradually add 2 cm depth of FA 10. Pour half of this mixture into another test - tube and place it in the almost boiling water bath for a few minutes for comparison (Blue/ bluish -green) FA 9 turned green upon adding FA 10. The resultant solution turned yellowish-green/ brighter green after warming. 3 To a 1 cm depth of FA 9 in a boiling-tube, gradually add 3 cm depth of aqueous sodium hydroxide. Filter the resultant mixture and collect the filtrate. Keep the filtrate for Test 4. Pale blue/ Blue -green/greenish-blue ppt formed is insoluble in excess NaOH(aq). The residue is dark blue. The filtrate is a colourless solution. 4 To a 1 cm depth of filtrate in a test -tube, add 1 cm depth of nitric acid, followed by AgNO3(aq). To a portion of the resultant mixture, add NH3(aq). White ppt formed with AgNO 3(aq) is soluble in aqueous ammonia to give a colourless solution. (b) (i) Cation Copper(II) ion/ Cu2+ (ii) Either one of the following: • Colour of the white ppt cannot be seen clearly in dark-coloured FA 9 solution. • Addition of NH 3(aq) could precipitate/form dark blue complex with Cu2+ if present. (c) (i) [CuCl4]2− In Test 2, when FA 10/ saturated NaY/Y− is added to FA 9, a yellow complex is formed. The solution is green due to presence of both blue and yellow complexes/ The solution turned green due to more yellow complex formed. (ii) The conversion of blue to yellow complex is an endothermic process. Upon warming, the formation of yellow complex is favoured/ POE of Equation 1 shift to the right to absorb some of the added heat.

River Valley High School Pg 23 of 20 JC 2 H2 Chemistry 9729 2022 Preliminary Examination Paper 4 4 (a) (i) H+(aq) + OH−(aq) → H2O(l) (ii) Amount of OH− in 25.0 cm3 = 25.002 1.00 1000 = 0.0500 mol Volume of HCl required for complete neutralisation = 0.0500 10001.50  = 33.33 cm3 Heat evolved = 57 000  0.0500 = (25.0 + 33.33)  4.18  T T = 11.7 C (b) Procedure 1. Fill a 50.00 cm3 burette to 0.00 cm3 mark with 1.50 mol dm−3 HCl(aq). 2. Place a Styrofoam cup in a 250 cm3 beaker to prevent it from toppling. 3. Pipette 25