RI Prelim P4 Answers

- 1 - 2018 Y6 H2 Chemistry Preliminary Exams Paper 4 (Suggested Solutions) 1(a) Volume of FA 2 needed = 5 30 x 250 = 41.67 cm3 1(b) Dilution of FA 2: Final burette reading / cm3 41.00 Initial burette reading / cm3 0.00 Volume of FA 2 used / cm3 41.00 Titration Results: Titration number 1 2 Final burette reading / cm3 24.80 24.80 Initial burette reading / cm3 0.00 0.00 Volume of FA 1 used / cm3 24.80 24.80 Values used   1(c) Average volume of FA 1 used = 24.80 + 24.80 2 = 24.80 cm3 25.0 cm3 of FA 4 produced an amount of iodine which required 24.80 cm3 of FA 1. 1(d) Amount of thiosulfate ions = 0.0500 × 24.80 1000 = 1.24 × 10−3 mol Mole ratio of Cu2+ : I2 : S2O32− = 2 : 1 : 2 Amount of Cu2+ in 25.0 cm3 of FA 4 = amount of S2O32− = 1.24 × 10−3 mol [Cu2+] in FA 4 = 1.24×10-3 25.0×10-3 = 0.0496 mol dm⁻3 1(e) [Cu2+] in FA 2 = 0.0496 x 250 41 = 0.302 mol dm⁻3 1(f) Amount of NaI in 10 cm3 of 50 g dm−3 sodium iodide = 50 23.0 + 126.9 x 0.010 = 0.00334 mol Amount of NaI required to react with Cu2+ in 25.0 cm3 of FA 4 = 2 x 25.0 x 10−3 x 0.0496 = 0.00248 mol When using 50 g dm −3 sodium iodide, amount of iodide is still in excess. Hence there is no effect on the titre volume as the same amount of I2 will be produced.

- 2 - 2(b)(i) Results: No V FA5/ cm3 VFA6/ cm3 T1/ oC T2/ oC ΔT/ oC 1 10.0 30.0 29.0 32.0 3.0 2 20.0 20.0 29.6 35.2 5.6 3 30.0 10.0 29.6 33.3 3.7 4 35.0 5.0 30.0 31.6 1.6 5 25.0 15.0 29.6 35.4 5.8 6 15.0 25.0 29.0 33.2 4.2 2(b)(ii) 2(b)(iii) maximum temperature change of reaction mixture, ΔTmax = 6.5 0C volume of FA 5 required for complete reaction, Vrxn = 23.5 cm3 2(c)(i) To obtain ΔTmax, VFA6 = 40.0 – 23.5 = 16.5 cm3 Amount of S2O32− = 16.5 1000 x 0.100 = 0.00165 mol Amount of NaClO in 23.5 cm3 of FA 5 = 4 x amount of S2O32− = 0.00660 mol Concentration of NaClO in FA 5 = 0.00660 0.0235 = 0.281 mol dm−3 2(c)(ii) Heat given out = 40.0 x 4.18 x 6.5 = 1086.8 J ΔH1 = − 1086.8 0.00165 J mol−1 = −659 kJ mol−1 2(d) q = mcΔTmax = n × ΔH1  ΔTmax = n  H1 m ൈ c When VFA5 and VFA6 are doubled, both n and m are doubled (ΔH1 and c remain constant). Hence ΔTmax is unaffected. 2(e)(i) Amount of S2O32− = 0.030 x 0.100 = 3.00 x 10−3 mol Amount of ClO− = 4 x amount of S2O32− = 0.0120 mol Mass of NaClO = 0.0120 x (23.0 + 35.5 + 16.0) = 0.894 g 1.0 2.0 3.0 4.0 5.0 6.0 7.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 ΔT/ oC VFA 5/ cm3 Vrxn ΔTmax

- 3 - 2(e)(ii) Assuming a percentage purity of 80 % NaClO, Mass of FA 7 required = 0.894 0.80 = 1.12 g 2(e)(iii) Using an analytical balance, weigh accurately about 0.400 g of FA 7. Record the mass of the weighing bottle and the mass of FA 7, m1. Using a burette, add 30.00 cm 3 of FA 6 in a clean and dry Styrofoam cup . Place the cup inside a second Styrofoam cup which is placed in a 250 cm 3 glass beaker to prevent it from tipping over. Place a thermometer into the cup containing FA 6. Stir gently, measure and record the initial temperature of FA 6, T1. Pour FA 7 from the weighing bottle into the cup containing FA 6. Using the thermometer, stir to dissolve FA 7, measure and record the highest temperature of the mixture, T2. Weigh and record the mass of the emptied weighing bottle, m2. Mass of FA 7 used = m1 − m2. Maximum temperature change = T2 − T1. Repeat the experiment using 0.600 g, 0.800 g, 1.200 g, 1.400 g and 1.600 g of FA 7. Plot graph of maximum temperature change against mass of FA 7 used. A graph similar to the following graph would be obtained: m 3 and ΔTmax may be obtained. Percentage purity of NaClO in FA 7 = 0.894 m3 x 100 % Heat change, q = 30.00 x 4.18 x ΔTmax Enthalpy change of reaction 2, ΔH2 = − 30.00 x 4.18 x Tmax amount of S2O3 2- = − 30.00 x 4.18 x Tmax 0.003 = − 41800 ΔTmax J mol−1 Temperature rise / oC Mass of FA 7 ΔTmax m3

- 4 - 2(e)(iii) 4NaClO(s) + Na2S2O3(aq) + 2NaOH(aq) ⎯⎯→ 4NaCl(aq) + 2Na2SO4(aq) + H2O(l) 4NaClO(aq) + Na 2S2O3(aq) + 2NaOH(aq) 3(a) Test Observations (i) Using a measuring cylinder, add 10 cm3 of FA 2 into a boiling tube. Add 4 spatulas of FA 8. Warm the mixture cautiously till boiling. Leave to cool for 5 minutes. Filter the mixture and keep the filtrate for tests (ii) and (iii). Blue FA 2 turns colourless. Red-brown/brown/black residue colourless filtrate (ii) To 1 cm depth of the filtrate, add aqueous ammonia. White ppt formed, soluble in excess NH3 to give a colourless solution. (iii) To another 1 cm depth of the filtrate, add 2 spatulas of solid ammonium chloride, followed by aqueous ammonia. No ppt formed. (v) Using a spatula, add a very small quantity of FA 8 to the boiling tube containing FA 10 solution from test (iv). Swirl the mixture gently and record your observations. Continue to add more FA 8 in small quantities with swirling, until no further colour change is observed. Record all colour changes observed. Yellow solution turns green Green solution turns blue Blue solution turns green Green solution turns violet/purple Effervescence of H2 gas extinguished lighted splint with a ‘pop sound. (vi) To 1 cm depth of the filtrate from test (v), add an equal volume of aqueous hydrogen peroxide. Violet solution turns red brown/brown/ orange/orange-brown. Effervescence of O2 gas relighted glowing splint. 3(b)(i) Identity of FA 8 Evidence Zn In test (i), Zn was oxidised to Zn 2+ by Cu 2+ in FA 2. In test (ii) The Zn 2+ formed a white ppt of Zn(OH) 2, soluble in excess NH 3 to give a colourless solution. 3(b)(ii) NH4Cl → NH4+ + Cl− NH3 + H2O ⇌ NH4+ + OH− ----(1) In the presence of NH4+ from the full dissociation of ammonium chloride, the dissociation of NH3 is suppressed (or position of equilibrium of (1) lies to the left). The concentration of OH− is too low for the ionic product to exceed Ksp or for ppt to form. 3(b)(iii) The grey ppt was V(OH)2. ΔH2 ΔH1 4ΔHsol ΔHsol = ¼ (ΔH2 − ΔH1)

- 5 - 3(c) Add FA 2 to the four unknown solutions. The solution that produces a bluish green ppt of CuCO 3 is Na2CO3. To the three remaining solutions that did not give a blue −green ppt, add an equal volume of Solution X and warm. The solution that produces a reddish brown ppt of Cu2O is CH3CH2CHO. The two remaining solutions that did not produce any ppt are Al2(SO4)3 and CH3COOH. To the two remaining solutions, add the unknown that was identified as Na 2CO3. The solution that produces effervescence of CO2 is CH3COOH. The solution that produces effervescence of CO2 and a white ppt of Al(OH)3 is Al2(SO4)3.