2025 RI H2Chem Prelims P4 Answers

-1- 2025 RI H2 Chemistry Paper 4 – Suggested Solutions 1(a)(i) tests observations Test FA 1 with Universal Indicator paper. Universal Indicator paper turns orange. pH is 3 Add 1 cm depth of FA 1 into a test- tube. Add aqueous sodium hydroxide, slowly with shaking, until no further change is seen. Filter the mixture into a clean test-tube. To the filtrate, add dilute nitric acid drop-wise until in excess. Off-white/white ppt formed, insoluble in excess NaOH. Off-white ppt turned brown on contact with air. Off-white/brown residue. Colourless filtrate. White ppt formed, soluble in excess dilute nitric acid to give a colourless solution. Add 1 cm depth of FA 2 into a test- tube. Add 1 cm depth of dilute nitric acid. Effervescence. CO2 gas evolved gave a white ppt with limewater. 1(a)(ii) identity evidence cation in FA 1 Mn2+ In test 2 , FA 1 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. cation in FA 1 Al3+ In test 2, FA 1 reacted with excess NaOH(aq) to form a colourless filtrate after filtration. On adding dilute nitric acid, a white ppt of Al(OH)3 is formed. Also, in test 1, FA 1 is acidic. anion in FA 2 CO32– In test 3 , FA 2 reacted with dilute nitric acid to give effervescence of CO2 gas which gave a white ppt with limewater.

-2- 1(b) test observations with FA 3 observations with FA 4 Add 1 cm depth of FA 3 to a test-tube. Add all of the magnesium turnings provided in the vial to this test-tube. Effervescence. Colourless H2 gas is evolved and ‘pops’ with a lighted splint. no observable change Add about 1 cm depth of FA solution to a test -tube. To this test-tube, add 8 drops of sodium hydroxide solution followed by iodine solution, dropwise, until a permanent orange/red colour is present. Solution FA 3 turns orange OR no yellow ppt Solution FA 4 turns orange. Yellow ppt forms. Add 1 cm depth of aqueous silver nitrate to a test -tube. Then slowly add 1 cm depth of aqueous sodium hydroxide. Add aqueous ammonia slowly, with shaking, until the precipitate just dissolves. You may use a clean glass rod to stir the mixture and help dissolve the precipitate. Add 1 cm depth of FA 4 to this mixture, shake the tube and place it in the test -tube rack to stand. no observable change Silver mirror formed 1(b)(iv) identity Y in FA 3 CH3COOH or ethanoic acid Z in FA 4 CH3CHO or ethanal 2(a) Total volume of FA 6 added, VFA 6 / cm3 Total volume of solution in the cup, Vtotal / cm3 Maximum temperature, T / oC T = T – T0 / oC (Vtotal × T) / cm3 oC 0.00 40.0 29.4 0.0 0.00 5.00 45.0 30.4 +1.0 45.0 10.00 50.0 31.0 +1.6 80.0 15.00 55.0 31.6 +2.2 121 20.00 60.0 32.2 +2.8 168 25.00 65.0 32.2 +2.8 182 30.00 70.0 31.8 +2.4 168 35.00 75.0 31.6 +2.2 165 40.00 80.0 31.4 +2.0 160 45.00 85.0 31.2 +1.8 153

-3- 2(b)(i) Veq = 22.0 cm3 Maximum value of (Vtotal × T) = 183 2(b)(ii) n(NaHCO3) used = 40.0 1000 x 0.6 = 0.0240 mol = n(NaOH) reacted concentration of NaOH in FA 6 = 0.024  22 1000 = 1.09 mol dm−3 2(b)(iii) From the graph, max (Vtotal × T) = 183 heat change, q = m x c x T = (m x T) x c = (Vtotal × T) x c = 183 x 4.18 = +764.94 J H1 = −(764.94  10−3)  0.0240 = −31.9 kJ mol−1 (Vtotal × T) / cm3 oC Total volume of FA 6 added, VFA 6 / cm3 0 20 40 60 80 100 120 140 160 180 200 220 0 5 10 15 20 25 30 35 40 45 50 183 22.0

-4- 2(b)(iv) (Vtotal × T) will vary more linearly with (or is directly proportional to) the total volume of FA 6 added while curves are obtained when p lotting T against total volume of FA 6 added. It is more accurate to extrapolate straight lines than curves. 2(c)(i) Taverage = (40.0 × 28.6) + (50.0 × 28.9) 40.0 + 50.0 = 28.77 C = 28.8 C (3 s.f.) 2(c)(ii) heat change, q = (40.0 + 50.0)(4.18)(26.2 – 28.8) = −978.12 J n(NaHCO3) used = 0.60  40.0 1000 = 0.0240 mol Since NaHCO3 is the limiting reagent, H2 = + (−978.12  10−3)  0.0240 = +40.8 kJ mol−1 2(d)(i) Let mass of NaHCO3 used be 5.0 g, n(NaHCO3) used = 5.0 / 84.1 = 0.059453 mol n(HCl) needed = 0.059453 mol volume of HCl needed = 0.059453 / 2.00 = 0.02973 dm3 = 29.7 cm3 Since HCl used must be in excess, use 50.0 cm3 of 2 mol dm−3 HCl. 2(d)(ii) 2.5 2(d)(iii) Tmax obtained by the graphical method is more accurate as heat gained from surroundings has been accounted for by extrapolation to find the lowest temperature reached from the graph. 2(d)(iv) Addition of HCl solution in small portions can minimise the acid spray caused by effervescence of CO2(g). temperature / oC time / min Tmax

-5- 3(a)(i) Dilution of FA 6 Final burette reading / cm3 33.50 Initial burette reading / cm3 21.00 Volume of FA 6 used / cm3 12.50 Titration results Final burette reading / cm3 29.10 39.10 Initial burette reading / cm3 10.00 20.00 Volume of FA 7 used / cm3 19.10 19.10 Values used (✓) ✓ ✓ 3(a)(ii) Average volume of FA 7 used = 19.10 + 19.10 2 = 19.10 cm3 3(b)(i) Concentration of KHC8H4O4 in FA 8 = 8.15 204.2 = 0.03991 mol dm−3 Amount of KHC8H4O4 in 25.0 cm3 of FA 8 = 0.03991  0.0250 = 9.978  10−4 mol Amount of NaOH used = 9.978  10−4 mol Concentration of NaOH in FA 7 = 9.978  10-4 0.01910 = 0.05224 mol dm−3 = 0.0522 mol dm−3 (3 s.f.) 3(b)(ii) Concentration of NaOH in FA 6 = 250.0 12.50 × 0.05224 = 1.04 mol dm−3 3(c)(i) percentage error = 1.09 - 1.04 1.04 × 100% = 4.81% 3(c)(ii) The volumetric titration method is more accurate as a burette is used to measure equivalence volume accurately. However, the thermometric titration method used a less accurate method of extrapolating the graph to determine Veq. OR There in greater inaccuracy in data obtained using the thermometric titration method due to heat loss to surroundings whereas volumetric titration eliminates the need to account for heat exchange with surroundings. 4(a) [NaClO3] in reaction mixture = (0.00144 x 30 1000) ÷ 90 1000 = 0.000480 mol dm–3 Limiting reagent is ClO3– Maximum [ClO2] = 0.000480 mol dm–3 Applying Beer-Lambert’s Law, A = cℓ Maximum absorbance value = (1250)(0.000480)(1) = 0.600

-6- 4(b) Procedure 1. Using a 50 cm 3 measuring cylinder, transfer 30.0 cm 3 of NaCl(aq) into a 250 cm3 conical flask (or beaker). 2. To the same conical flask , add 30.0 cm 3 of H 2SO4(aq) using another 50 cm3 measuring cylinder. 3. Measure 30.0 cm3 of NaClO3(aq) using a 50 cm3 measuring cylinder. 4. Add the NaClO3(aq) into the same 250 cm 3 conical flask in step 1. Start the stopwatch immediately. Swirl the conical flask to ensure even mixing. 5. Using a 10 cm 3 pipette, transfer 10.0 cm 3 of the reaction mixture (aliquot) into a 50 cm3 beaker. 6. At time 2 min, place the beaker containing the aliquot in an ice bath and allow it to cool for 3 minutes. 7. Measure the absorbance of the aliquot using the UV-vis spectrometer. 8. Repeat steps 5 - 7 at 6, 10, 14, 18, 22, 26 minutes. 4(c) 1st t1/2 = 2nd t1/2, t1/2 is constant and reaction is first order with respect to ClO3–.