RVHS Prelim H2 Chemistry P4 Soln

River Valley High School 9729/02/PRELIMS/25 2025 Preliminary Examination [Turn over River Valley High School 9729 H2 Chemistry Preliminary Examination Paper 4 Suggested solution 1 (a) (i) Results Titration number 1 2 Final burette reading /cm3 23.75 23.75 Initial burette reading /cm3 0.00 0.00 VFA2 /cm3 23.75 23.75 Subject supervisor’s reading: 23.75 cm3 2 marks if difference  0.20 cm3, 1 mark if difference  0.30 cm3 [5] (ii) average volume of FA 2 used = 23.75 23.75 2 + = 23.75 cm3 [1] (iii) [Ag+(aq)] = 23.75 100.011000 1000 = 0.02375 = 0.0238 mol dm–3 [1] (iv) Fe2+(aq) + Ag+(aq) = Fe3+(aq) + Ag(s) I/ mol dm−3 (0.1)/2 (0.1)/2 0 - C -x -x +x E 0.02375 0.02375 x Initial [Fe2+] or initial [Ag+] = 0.1 / 2 = 0.05 mol dm–3 [Fe3+(aq)] = 0.05 – 0.02375 = 0.02625 = 0.0263 mol dm–3 OR Initial amount of Fe2+ or initial amount of Ag+ = 0.1 × 100 1000 = 0.01 mol [2]

2 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination [Fe3+] = 2000.01 0.023751000 200 1000 − = 0.02625 = 0.0263 mol dm–3 (v) Kc = 2 0.0263 (0.0237) = 46.8 mol–1 dm3 [2] (b) Disagree. Either of the following: • The equilibrium is established in 4 hours and thus the position of equilibrium does not shift within minutes of titration. • Titration is carried out very quickly such that it will not shift the position of equilibrium/ no time to establish new equilibrium. • The value of Kc implies that position of equilibrium lies far right, the impact of titration does not affect position significantly. • FA 1 is left to stand for 4h, all Ag would have aggregated as solid at the bottom, which was left undisturbed. There is no Ag in the aliquot, so backward reaction will not occur, shifting of position of equilibrium will not occur. [1] (c) Use a lower concentration of FA1/ FA2 , increase volume used, lower percentage uncertainty. Use higher volume of FA 1 used, lower percentage uncertainty of Kc. Conduct the experiment in a thermostatically controlled water bath OR controlled at constant temperature. Filter the equilibrium mixture prior to titration to remove the solid so that the backward reaction will not occur during titration to affect the titre and value of Kc. [1] [Total: 13]

3 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination [Turn over 2 (a) Results V/cm3 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 T/ °C 31.0 35.0 38.0 40.0 41.5 42.0 41.0 40.0 39.0 (b) (i) [3] (ii) volume of FA 4 used = 23.00 …………………cm3 [1] (iii) Energy change = (25.0 + 23.0) (4.18) (42.20 – 31.0) = 2247.2 J = 2250 J [1] 23.00 42.20

4 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination (iv) Amount of Ba(OH)2 = 0.95 (0.025) = 0.02375 = 0.0238 mol [1] (v) ∆Hneut = (–2247.2J / 1000) / 2(0.02375) = – 47.3 kJmol−1 [1] [1] Accuracy −43 ≤ ∆Hneut ≤ −53 [2] (vi) [CHX(COOH)2] = 0.02375 mol / 0.023 = 1.033 mol dm–3 Molar mass = 141.3 / 1.033 = 136.8 g mol−1 Mr = 137 (3sf) [1] (vii) Acid is CHCl(COOH)2 [1] (viii) percentage error = (138.5–136.8 ) / 138.5 ×100% = 1.20% [1] M2: 3sf for Q1 (except (a)(ii) and Q2 (except (b)(ii)) M3: units for Q1 and Q2 (except (b)(vi)) [3] (c) In a conventional titration, the V neut was determined directly by a colour change which produces a sharper and more reliable end-point. In a thermometric titration, V neut was determined indirectly from the maximum temperature rise determined graphically, and subject to errors e.g. heat loss. Improvement: V∆T should be used to determine the Vneut, which would have taken into account the increasing volume of the reaction system. [1] [Total: 16]

5 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination [Turn over 3 (a) Table 3.1 test observations (i) To a 1cm depth of FA 5 in a test-tube add a 1 cm depth of sodium hydroxide, FA 5 is colourless. With NaOH: Pale or light brown/ off white ppt formed and insoluble in excess . Ppt rapidly turn brown (on contact with air). then add several drops of hydrogen peroxide. With H2O2, Brown ppt turned dark brown/ black. Effervescence. Colourless, odourless gas evolved that relight a glowing splint. Gas is oxygen. (ii) To a 1 cm depth of FA 6 in a test tube add aqueous sodium hydroxide. FA 6 contains yellow solution with orange ppt. Green ppt is insoluble in excess. Green ppt turned brown on contact with air (iii) To a 1 cm depth of FA 6 in a test-tube add several drops of hydrogen peroxide and FA 6 turned brown Effervescence. Colourless, odourless gas evolved that relight a glowing splint. Gas is oxygen. then add aqueous sodium hydroxide. Red-brown ppt formed insoluble in excess. (iv) To a 1 cm depth of FA 6 in a test-tube add a 1 cm depth of sulfuric acid and then add a few drops of FA 7. Purple FA 7 turned colourless/ decolourised Allow purple to (pale) yellow/ pale orange (v) To a 1 cm depth of aqueous potassium iodide in a test-tube add a few drops of FA 7 With KI, Brown/ black ppt in brown solution.

6 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination then add a few drops of aqueous starch. With starch, Brown solution turned dark blue/ blue - black [5] (b) FA 5 FA 6 FA 7 Metal ions Mn2+/ Mn(II) Fe2+/ Fe(II) MnO4−/ Mn(VII) [3] (c) (i) Test Observation Add AgNO3(aq) No observable change/ no ppt/ solution remain (pale) yellow/ orange Add Ba(NO 3)2 (aq), followed by nitric acid White ppt formed with Ba2+ insoluble in HNO3 / White ppt formed with Ba 2+, no gas evolved [2] OR Test Observation Add HNO3(aq) No gas evolved/ no observable change Add Ba(NO3)2 (aq) White ppt formed. (ii) Anion Evidence SO42− (White ppt of Ag 2SO4) White ppt of BaSO4 formed which is insoluble in nitric acid. Or SO42− No SO2 evolved with HNO3. White ppt of BaSO4. [Total: 12]

7 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination [Turn over 4 (a) In case some of the light is absorbed/ blocked (allow reflect) by fingerprints / dirt/ oil/ contaminant. [1] (b) [2] (c) (i) Mass of CV = 0.25 × 1.80 × 10−3 × (407.5) = 0.18338 =0.183 g [1] (ii) Mass of crystal violet (0.0018 g) is too small to be measured directly. Percentage uncertainty/ error of mass measurement is (too) big. [1] (d) (i) 1. Using a 10 cm 3 pipette, add 10 cm 3 of 1.80 × 10 −5 mol dm−3 crystal violet into a clean and dry 250cm3/ 100 cm 3 conical flask/ 100 cm3 beaker. 2. Using a 10 cm 3 measuring cylinder, add 10 cm 3 of 0.200 mol dm−3 NaOH to the conical flask. Start the stopwatch immediately. Swirl the reaction mixture. 3. At 20s, using a dropper, quickly transfer some of the mixture from the conical flask to a cuvette and place the cuvette in the colorimeter. 4. Measure the absorbance of the mixture in a clean and dry cuvette at 565 nm every 30s until the solution decolourises. 5. Plot ln [A] against t, gradient = −k’ 6. Using a burette, add 8.00 cm3 of NaOH into a 10 cm3 volumetric flask. Make to the mark using deionised water, adding dropwise near the mark. Shake to ensure a homogeneous solution. 7. Repeat step 1. 8. Transfer the diluted NaOH(aq) from the volumetric flask into a clean and dry 100 cm 3 conical flask directly. Start the stopwatch immediately. Swirl the reaction mixture. 9. Repeat steps 3 to 5.

8 River Valley High School 9729/04/PRELIMS/25 2025 Preliminary Examination 10. Repeat step 6 - 9 using 6.00 cm 3, 4.00 cm 3 and 2.00 cm 3 of NaOH, topping up to the 10 cm 3 mark using deionised water each time. (ii) For each experiment, calculate [OH−]rxn mixture = NaOH total V 0.2 20V or 1000  Since k’ of each experiment = gradient of ln A against t (Choos