2024 RVHS H2 Chemistry Prelim P4 (Ans)

2024 JC 2 H2 Chemistry 9729 Prelim Exam Paper 4 Suggested Solutions 1 (a) Titration number 1 2 Final burette reading /cm3 25.05 25.15 Initial burette reading /cm3 0.00 0.00 Volume of FA 2 (added) /cm3 25.05 25.15 (b) average volume of FA 2 used = 25.05 25.15 2 + = 25.10 cm3 volume of FA 2 used = ………25.10 cm3……….. (c) (i) amount of MnO4− used = 25.10 1000  0.010 = 2.51  10−4 mol amount of MnO4− = ……2.51  10−4 mol…….. (ii) amount of Fe2+ in 25.0 cm3 of FA 4 = 2.51  10−4  5 = 1.255  10−3 = 1.26  10−3 mol amount of Fe2+ = ……1.26  10−3 mol…….. (iii) amount of Fe2+ in 250 cm3 of FA 4 = 1.255  10−3  250 25.0 = 1.255  10−2 mol concentration of Fe2+ in FA 1 = 21.255 10 25.0 1000 − = 0.502 mol dm−3 concentration of Fe2+ = ……0.502 mol dm−3….... (iv) Mass of Fe(NO3)2 in 1 dm3 of FA 1 = 0.502  (55.8 + 2(14.0) + 6(16.0)) = 90.26 g Percentage by mass of Fe(NO3)2 = 90.26 125.4  100% = 72.0% percentage by mass of Fe(NO3)2 = ………72.0% …………..

River Valley High School Pg 2 of 9 JC 2 H2 Chemistry 9729 2024 Preliminary Examination Paper 4 Suggested Solutions (d) The titre value will be lower, because some of the Fe2+ in FA 1 would have been oxidised to Fe3+. Hence, less Fe2+ would be available for titration with KMnO4. Note: Sulfuric acid is added to prevent the oxidation of Fe2+ to Fe3+. (e) (i) In the titration of Fe 2+(aq) with MnO 4−(aq) in acidic medium, yellow Fe 3+(aq) and colourless Mn2+(aq) are formed. The end-point is reached when one excess drop of purple MnO4−(aq) is added, turning the solution from yellow to orange. (ii) There is no distinct/ visible/ significant colour change at the end-point. Hence, it would be impossible to know when the reaction is completed. OR H2O2 undergoes disproportionation/ decomposition, hence a higher volume of titrant will required. OR Fe2+/ Fe3+ present in the solution catalyses the decomposition of H2O2. (f) (i) % uncertainty in pipette = 0.06 25.0  100% = 0.240% % uncertainty in burette = 0.05 0.05 29.70 +  100% = 0.337% % uncertainty in measuring cylinder = 0.25 25.00  100% = 1.00% (ii) Burette Note: Measuring cylinder is not considered as FA 3 is added in excess.

River Valley High School Pg 3 of 9 JC 2 H2 Chemistry 9729 2024 Preliminary Examination Paper 4 Suggested Solutions 2 (a) Expt VFA 6 /cm3 𝑉H2O /cm3 Reaction time, t /s lg(VFA 6) lg (rate) 1 20.00 0.00 17.8 1.30 1.75 2 10.00 10.00 63.9 1.00 1.19 3 13.00 7.00 39.1 1.11 1.41 4 17.00 3.00 24.6 1.23 1.61 (b) (c) (i) Gradient of line = 1.63 1.20 1.235 1.005 − − = 1.87 (3 s.f.) m = 2 (nearest integer) gradient = ………1.87…………….. m = ……….2………………..

River Valley High School Pg 4 of 9 JC 2 H2 Chemistry 9729 2024 Preliminary Examination Paper 4 Suggested Solutions (ii) From graph, when lg(12.30) = 1.09, lg( 1000 reaction time ) = 1.36 reaction time = 1.36 1000 10 = 43.7 s reaction time = ………43.7 s……….. (d) (i) expt volume of FA 5 /cm3 volume of FA 6 /cm3 volume of FA 7 /cm3 volume of water /cm3 reaction time /s 2 10 10 20 10 t2 5 5 10 20 15 t5 (ii) Halving the concentration (or volume) of Fe3+ will cause the rate to be halved. Since rate is inversely proportional to reaction time, time taken for experiment 5 will be double that of experiment 2, i.e. t5 = 2t2. (iii) rate = k [Fe3+][I–]2 (iv) Kc for step 1 = I I 2+ 3+ - [Fe ] [Fe ][ ] Since FeI2+ is an intermediate, rate = k [FeI2+][I–] = kKc [Fe3+][I–]2 = k’ [Fe3+][I–]2. This matches the rate equation for the reaction, hence the proposed mechanism is consistent with the observed kinetics data. (e) Disagree with student. The reaction time should be shorter since less/ smaller amount of thiosulfate ions is present to react with iodine.

River Valley High School Pg 5 of 9 JC 2 H2 Chemistry 9729 2024 Preliminary Examination Paper 4 Suggested Solutions 3 (a) Tests Observations 1. Test the FA 9 solution using Universal Indicator paper. FA 9 is an (dark) orange solution. [✓] UI paper turns orange. [✓] FA 9 has a pH of 2/3. [✓] 2. To a 2 cm depth of FA 9 in a test tube, add a 2 cm depth of FA 10 and shake the mixture thoroughly. Observe the mixture until there is no further change. Effervescence observed. [✓] Colourless and odourless gas evolved. Gas evolved rekindled a glowing splint. [✓] O2(g) evolved. [✓] Test tube felt warm to touch. [✓] 3. To a 2 cm depth of FA 9 in a boiling tube, add aqueous sodium hydroxide until there is no further change. Warm the mixture obtained. Red-brown ppt formed. Ppt was insoluble in excess NaOH(aq). [✓] Colourless, pungent gas evolved. Gas evolved turned moist red litmus blue. [✓] NH3(g) evolved. [✓] 4. To a 2 cm depth of FA 9 in a test tube, add aqueous sodium carbonate. (Red-/ orange-) brown ppt formed. [✓] Bubbles of gas/ effervescence observed. [✓] Colourless, odourless gas evolved. Gas evolved formed white ppt with limewater. [✓] CO2(g) evolved. [✓] 5. To a 2 cm depth of FA 9 in a test tube, add aqueous potassium thiocyanate, KSCN(aq). Blood red/ deep red/ dark red colouration observed. [✓] 6. To a 2 cm depth of FA 9 in a test tube, add aqueous barium nitrate. Leave the mixture to stand for 5 minutes and discard the liquid. To the remaining solid, add dilute nitric acid. White ppt formed. Ppt was insoluble in dilute nitric acid. [✓]

River Valley High School Pg 6 of 9 JC 2 H2 Chemistry 9729 2024 Preliminary Examination Paper 4 Suggested Solutions (b) Fe3+. Fe3+ undergoes hydrolysis and forms H3O+, resulting in an acidic solution. (c) Identity of ions Supporting evidence NH4+ In test 3, colourless and pungent NH3(g) evolved that turned moist red litmus blue. SO42− In test 6, white ppt BaSO 4(s) formed that was insoluble in dilute nitric acid. (d) Ligand exchange reaction (e) Test: To 2 cm depth of FA 10 in a test tube, add acidified KMnO4(aq). Observations: Purple KMnO 4(aq) decolourises. Colourless, odourless gas evolved that rekindled a glowing splint. Explanation: Ecell = +1.52 − (+0.68) = +0.84 V > 0, hence reaction between MnO 4− and H2O2 is spontaneous. OR Test: To 2 cm depth of FA 10 in a test tube, add acidified KI(aq). Observations: Colourless KI(aq) turns brown/ black deposits formed. Explanation: Ecell = +1.77 − (+0.54) = +1.23 V > 0, hence reaction between H2O2 and KI is spontaneous. OR Test: To 2 cm depth of FA 10 in a test tube, add dilute HCl. Observations: Greenish-yellow, pungent gas evolved that turned moist blue litmus red then bleached. Explanation: Ecell = +1.77 − (+1.36) = +0.41 V > 0, hence reaction between H2O2 and HCl is spontaneous.

River Valley High School Pg 7 of 9 JC 2 H2 Chemistry 9729 2024 Preliminary Examination Paper 4 Suggested Solutions 4 (a) NaOH(aq) + 1 x HxA(aq) → 1 x NaxA(aq) + H2O(l) (b) Since varying volumes of FA 11 and FA 12 should be used for the series of experiments, the following combination of volumes can be chosen for the experiments: Experiment 1 2 3 4 5 6 VFA 11/ cm3 10 20 25 30 40 50 VFA 12/ cm3 50 40 35 30 20 10 The total volume of the mixture should be kept constant for all experiments. M45: suitability of volumes • minimum 6 experiments • total volume of mixture kept constant • do not award if student fixes volume of either FA 11 or FA 12 Procedure: 1. Using a 50 cm 3 measuring cylinder/ burette , measure 10.0 cm 3 of FA 11 into a styrofoam cup supported in a 250 cm3 beaker. 2. Using a thermometer, measure and record the initial temperature