RI 2024 Prelims P4 Solutions

-1- 2024 RI H2 Chemistry Paper 4 – Suggested Solutions 1(a)(i) Table 1.1 temperature of room temperature water, T1 / °C 30.4 temperature of hot water, T2 / °C 70.1 maximum temperature of combined water, T3 / °C 48.2 1(a)(ii) Heat energy lost by the hot water Heat energy gained by the room temperature = mc∆T water = mc∆T = 50 × 4.18 × (T2 – T3) = 50 × 4.18 × (T3 – T1) = 50 × 4.18 × (70.1 – 48.2) = 50 × 4.18 × (48.2 – 30.4) = 4577.1 J = 3720.2 J = 4580 J (3 s.f.) = 3720 J (3 s.f.) 1(a)(iii) Heat energy absorbed by the calorimeter Heat capacity of calorimeter, Ccal = heat energy lost by hot water = 856.9 T3 – T1 = 856.9 48.2 – 30.4 – heat energy gained by room temp water = 48.14 J oC−1 = 4577.1 – 3720.2 = 48.1 J oC−1 (3 s.f.) = 856.9 J = 857 J (3 s.f.) 1(b)(i) mass of capped weighing bottle and FA 1 / g 8.782 mass of capped weighing bottle after emptying FA 1 / g 4.745 mass of FA 1 used / g 4.037 time / min temperature / °C 0.0 30.4 0.5 30.4 1.0 30.4 1.5 30.4 2.0 30.4 2.5  3.0 36.2 3.5 36.4 4.0 36.5 4.5 36.3 5.0 36.2 5.5 36.1 6.0 36.0 6.5 36.0 7.0 35.9 7.5 35.8 8.0 35.6

-2- 1(b)(ii) 1(b)(iii) Tmin = 30.4 oC Tmax = 36.8 oC ∆T = +6.4 oC 1(b)(iv) Total heat change, q = heat absorbed by solution + heat absorbed by calorimeter = (mc∆T) + (Ccal ∆T) = (100 × 4.18 × 6.4) + (48.14 × 6.4) = +2983 J = +2980 J (3 s.f.) 1(b)(v) Amount of MgSO4 used = 4.037 24.3 + 32.1 + 4(16.0) = 0.03353 mol ∆Hsol = – q n(MgSO4) = – +2983 0.03353 = –89000 J mol−1 (or –89.0 kJ mol−1) (3 s.f.)

-3- 1(c) The heat energy absorbed by and heat capacity of calorimeter obtained in (a)(iii) is higher than expected. Hence, the ∆Hsol of FA 1 obtained in (b)(v) is more exothermic than expected (or the magnitude of ∆ Hsol of FA 1 obtained in (b)(v) is higher than expected). 2(a)(i) tests observations 1 Add 1 cm depth of FA 2 into a clean test-tube. To this test -tube add 10 drops of sodium hydroxide followed by iodine solution, dropwise, until a permanent orange colour is present. Warm the test -tube in the beaker of hot water for 2 minutes. (pale) Yellow ppt formed. 2 Add 1 cm depth of FA 2 into a clean test-tube. Add 8 drops of Fehling’s solution. Warm the test -tube in the beaker of hot water for 3 minutes. Brick-red ppt formed. 3 Test solution FA 2 with Universal Indicator paper. Universal Indicator paper turned dark yellow-green. pH is 7. 2(a)(ii) C C H O C O H H H H H 2(b)(i) tests observations 1 Using a 10 cm 3 measuring cylinder, add 2 cm3 of FA 3 into a clean boiling tube. Using another 10 cm3 measuring cylinder, measure out 7 cm 3 of aqueous sodium hydroxide. Slowly with shaking, add this completely to FA 3. Stir the contents of the boiling tube with a glass rod. Filter the mixture into a clean test -tube. The filtrate will be used for test 4. While waiting, proceed to test 2 and 3. Off-white ppt. rapidly turned brown on contact with air. Ppt. is insoluble in excess NaOH(aq). Brown residue. Colourless filtrate.

-4- 2(b)(i) tests observations 2 Add 1 cm depth of FA 3 into a clean test-tube. Add 2 cm depth of aqueous sodium carbonate. Off-white/ white ppt. Effervescence was observed. CO2 gas evolved gave a white ppt. with limewater. 3 Add 1 cm depth of FA 3 into a clean test-tube. Add 1 cm depth of aqueous silver nitrate. Filter the mixture and discard the filtrate. Wash the residue by pouring deionised water through it. Discard the washings. Place the filter funnel containing the residue into a test -tube containing 1 cm depth of dilute nitric acid. Carefully add aqueous ammonia to the filter funnel until it covers the residue. The filtrate will collect in the test-tube containing the dilute nitric acid. White ppt formed. White residue obtained. Colourless filtrate. White ppt formed. 4 Add 1 cm depth of the filtrate from test 1 into a clean test-tube. Add dilute sulfuric acid drop- wise, until in excess. White ppt. formed, soluble in excess dilute H 2SO4 to give a colourless solution. 2(b)(ii) cation evidence Mn2+ In test 1, FA 3 reacted with NaOH(aq) to give an off-white ppt. of Mn(OH)2 which was insoluble in excess NaOH(aq). On contact with air, Mn(OH)2 was oxidised to brown Mn(OH)3. Al3+ In test 2, FA3 reacted with Na2CO3(aq) to give a white ppt. of Al(OH)3 together with the effervescence of CO2 gas. anion evidence Cl− In test 3, FA 3 reacted with AgNO3(aq) to form a white ppt. of AgC l. White ppt of AgC l soluble in excess ammonia to form colourless solution of [Ag(NH3)2]+.

-5- 3(a)(i) Titration number 1 2 Final burette reading / cm3 21.90 41.80 Initial burette reading / cm3 0.00 20.00 Volume of FA 4 used / cm3 21.90 21.80 Values used ()   3(a)(ii) Average volume of FA 4 used = 21.90 + 21.80 2 = 21.85 cm3 3(b)(i) Amount of Na2S2O3 = 21.85 1000 × 0.100 = 2.185 × 10−3 mol Amount of I2 produced = (½) (2.185 × 10−3) = 1.093 × 10−3 mol = 1.09 × 10−3 mol 3(b)(ii) Amount of KMnO4 reacted = (2/5) (1.093 × 10−3) = 4.372 × 10−4 mol = 4.37 × 10−4 mol 3(b)(iii) Amount of KMnO4 in 25.0 cm3 FA 7 = 4.372 × 10−4 mol Amount of KMnO4 in 250 cm3 FA 7 = Amount of KMnO4 in 34.50 cm3 P = 4.372 × 10−4 × 10 = 4.372 × 10−3 mol Conc. of KMnO4 in P = 4.372 ×10-3 34.5×10-3 = 0.1267 mol dm−3 = 0.127 mol dm−3 (3sf) 3(b)(iv) Mass of KMnO4 = 0.1267 × (39.1 + 54.9 + 4×16.0) = 20.02 g = 20.0 g (3sf) 3(c)(i) Percentage = 21.0 − 20.0 21.0 ×100 = 4.76% (3sf) 3(c)(ii) I do not agree with the student, as KI used was in excess , hence the precision of the apparatus used is not relevant. 4(a) Excess H2NCH2CH2NH2(aq) must be used so that the equilibrium position of equation 4 lies to the right and the reaction mixture contains mainly [Ni(en)3]2+(aq). OR Excess H2NCH2CH2NH2(aq) must be used to ensure complete ligand exchange. 4(b) Dilution of FA 8, 2.00 x 10–3 mol dm−3 [Ni(H2O)6]2+(aq) 1. Using a burette, transfer 50.00 cm3 of FA 8 to a 100 cm3 volumetric flask. 2. Top up to the mark with deionised water, stopper the volumetric flask and shake this solution to obtain a homogeneous solution. Label this solution as solution 1. 3. Repeat steps 1 to 2 using the volumes of FA 8 shown in the table below to prepare solution 2 to solution 5. Solution Volume of FA 8 / cm3 [Ni(H2O)6]2+ / mol dm−3 1 50.00 1.00 x 10–3 2 40.00 8.00 x 10–4 3 30.00 6.00 x 10–4 4 20.00 4.00 x 10–4 5 10.00 2.00 x 10–4

-6- 4(c) Procedure 1. Using separate 25.0 cm 3 pipettes, transfer 25.0 cm 3 of 2.00 x 10 –3 mol dm–3 FA 8 and 25.0 cm3 of 6.00 x 10–3 mol dm–3 FA 9 into a 100 cm3 conical flask/ beaker. 2. Shake/Swirl the conical flask/ beaker to ensure a homogeneous solution. 3. Place the conical flask/ beaker in a thermostatically controlled water bath maintained at 80 °C for about 5 min. 4. Measure and record the temperature of the solution using a thermometer. 5. Remove the solution in conical flask/ beaker from the water bath and use the spectrometer to immediately measure and record the absorbance. 6. Repeat steps 3 – 5 at 70 °C, 60 °C, 50 °C and 40 °C. 7. For each absorbance obtained, read off the calibration line to determine the corresponding concentration of [Ni(en)3]2+(aq) at each temperature. 8. Calculate the Kc for each temperature according to the following method. Let the concentration of [Ni(en)3]2+(aq) obtained from the calibration line be x mol dm–3 [Ni(H2O)6]2+ + 3H2NCH2CH2NH2 ⇌ [Ni(en)3]2+ + 6H2O(l) initial / mol dm–3 2.00×10-3 2 = 1.00 × 10–3 6.00×10-3 2 = 3.00 × 10–3 0 − change / mol dm–3 – x – 3x + x − eqm /