NJC Prelim P4 Solutions

Suggested solution for 2017 SH 2 Prelim Practical 1 (a) (ii) Table 1: Weighings of FA1 Mass of weighing bottle and FA1 / g 3.696 Mass of weighing bottle and residual FA1 / g 3.497 Mass of FA1 transferred / g 0.199 Table 2: Titration of FA3 with FA4 using screened methyl orange as indicator 1 2 Final burette reading / cm3 30.00 36.90 Initial burette reading / cm3 9.00 16.00 Volume of FA4 used / cm3 21.00 20.90 (iii) average volume = 21.00 + 20.90 2 = 20.95 cm3 (b) (i) amount of HNO3 in 25.0 cm3 = amount of NaOH = 20.95 / 1000 x (9.10/40.0) = 4.77 x 10‒3 amount of HNO3 in 250.0 cm3 = 4.77 x 10‒2 (ii) amount of HNO3 in 25.0 cm3 = 25/1000 x 2 = 0.05 mol (iii) amount of HNO3 = 0.05 ‒ 4.77 x 10‒2 = 2.34 x 10‒3 (iv) amount of MgCO3.xH2O = 2.34 x 10‒3 / 2 = 1.17 x 10‒3 Mr = 0.208 / 1.15 x 10‒3 = 170.2 x = [180.7 – (24.3 + 12.0 + 16.0 x 3)] / 18 = 4.77 ≈ 5 (c) The acid will not be diluted and close to 2.00 mol dm−3, the titre value required to neutralise 25 cm3 of FA 3 will be about 250 cm3 / more than 10 times greater. Since this exceeds the capacity of burette, it is unsuitable.

2 (a) Table 1: Weighings of FA1 Mass of weighing bottle and FA1 / g 4.495 Mass of weighing bottle and residual FA1 / g 3.478 Mass of FA1 transferred / g 1.017 Table 2: Measurement of temperature Ti / oC 30.3 Ti / oC 33.3 T / oC 3.0 (i) Δqreaction3 = −mcΔT = − 40 x 4.18 x 3.0 = −502 J (iii) Hreaction1 Mg(s) + C(s) + (3+x)/2 O2(g) + xH2(g)  MgCO3.xH2O(s) + 2HNO3(aq) ‒490 −86.0 + 2HNO3(aq) −393.5 + 6 x −285.5 Mg(NO3)2(aq) + C(s) + (3+x)/2 O2(g) + (x+1)H2(g) Mg(NO3)2(aq) + CO2(g) + (x+1)H2O(l) By Hess’s law, Hreaction1 = ‒490 −393.5 + 6 x −285.5 – (−86.0) = −2510 kJ mol−1 OR If they use x = 3 By Hess’s law, Hreaction1 = ‒490 −393.5 + 4 x −285.5 – (−68.3) = −1957.2 kJ mol−1 = −1960 kJ mol−1 (c) (i) percentage error = (+ 0.1 x 2) / 3.0 x 100 = + 6.67 %

(ii) Hreaction3 = −502 / (1.017 / 174.3) = −86.0 kJmol−1 OR If they use x = 3 Hreaction3 = −502 / (1.017 / 138.3) = −68.3 kJmol−1 (ii) Error: The experiment did not take into account heat exchange with surrounding / calorimeter / heat loss to surrounding / calorimeter. Improvement 1: Use cooling curve method where temperature changes after mixing are measured at 30s intervals for about 3 min, then max temperature is extrapolated to time of mixing . This accounts for heat exchange with surrounding, hence a more accurate ∆T is obtained. Improvement 2: Calibration of the calorimeter by making use of reaction with accurately known enthalpy change. The calibrated heat capacity, C of the calorimeter takes into account heat exchange with surrounding, hence a more accurate ∆T is obtained. OR Error: The experiment did not repeat experiment to ensure consistent results. Improvement: Repeat experiment until T/m between a pair of experiments is within 5%. (d) Suggested solution: Procedure: 1. Using three 10.0 cm3 measuring cylinders, measure 5.0 cm3 of FB 1, FB 2 and FB 3. 2. Mix separately, 5.0 cm3 of FB 1 to 5.0 cm3 of FB 2 and 5.0 cm3 of FB 1 to 5 cm3 of FB 3, into 2 test-tubes 3. Measure for the rise in temperature using a 0.2 oC division thermometer Deduction:  If both reaction mixture gives a temperature rise, FB 1 is HCl.  If one of the 2 reaction mixture does not give a temperature rise, then this pair of solutions must be NH3(aq) and KOH(aq). HCl can be identified. Upon identifying HCl, 4. Using a 10.0 cm 3 measuring cylinder, add 10.0 cm 3 of HCl to a Styrofoam cup and measure the initial temperature, T 1, using a a 0.2 oC division thermometer.

5. Using another 10.0 cm 3 measuring cylinder, transfer 10 cm 3 of one of the bases into another Styrofoam cup. 6. Wash and dry the thermometer 7. Measure the initial temperature, T2, of the base solution. 8. Add HCl to base and stir with the thermometer and note the highest temperature reached, Tf and determine T. 9. Wash and dry the Styrofoam cup. 10. Repeat steps 4 – 9 but replace base with the other base solution. Assuming if FB 1 is HCl FB 2 FB 3 Ti / oC Tf / oC T / oC Deduction of results: Since HCl is 2 mol dm −3, the limiting reagent is the base and the same amount of base is used / same amount of H2O is produced in each reaction. When total volume of reaction mixture is kept constant and the no of moles of water formed in the reaction is the same, T depends only on the strength of the base used. As weak acid dissociates partially, a portion of the energy released from neutralisation is used to complete the dissociation of the weak acid. Therefore less energy is released compared to a neutralisation between a strong acid and a strong base. The rea ction mixture that gives a lower T must contain 1 mol dm −3 aqueous ammonia.

Alternative solution Procedure: 1. To 1 test-tube, add 5.0 cm3 of FB 1 using a 10.0 cm3 measuring cylinder and measure the initial temperature, T 1, using a 0.2 oC division thermometer. 2. Using another 10.0 cm3 measuring cylinder, measure 5.0 cm3 of FB 2. 3. Wash and dry the thermometer. 4. Measure the initial temperature, T2, of FB 2. 5. Transfer FB 2 into the test -tube. Stir with the thermometer and note the highest temperature reached, Tf and determine T. 6. Wash and dry the test-tube. 7. Repeat steps 1 – 6 but by mixing 5.0 cm3 of FB 1 to 5 cm3 of FB 3 and 5.0 cm3 of FB 2 to 5 cm3 of FB 3 respectively. FB 1 FB 2 FB 3 FB 1 T1 = T2 = Tf = T = T1 = T2 = Tf = T = FB 2 T1 = T2 = Tf = T = Results analysis: similar to first solution of using T to identify the 3 solutions. 3 (a) Expt Vol of FA 8 / cm3 Vol of DI / cm3 time / sec lg VFA 8 lg 1/t 1 20.00 0.0 46 1.30 -1.66 2 10.00 10.0 116 1.00 -2.06 3 15.00 5.0 64 1.18 -1.81 4 12.00 8.0 87 1.08 -1.94 5 8.00 12.0 251 0.903 -2.40

(ii) Order of reaction wrt Fe3+ = gradient of graph = (-1.725 ‒ -2.10) / (1.25 – 0.93) = 1.17 ≈ 1

4 Test Procedure Observations (a) To 6 cm 3 of X, add barium nitrate solution until in excess. White ppt. formed, insoluble in excess. (b) Filter the mixture from (a). Wash and retain the residue for test (c). Collect the filtrate for test (d) and (e). (c) To separate portions of the residue (i) add 2 cm 3 of hydrochloric acid A colourless and odourless gas evolved, which forms a white ppt with Ca(OH)2 (ii) add 2 cm 3 of organic compound Z [You are to test for any gas evolved] A colourless and odourless gas evolved, which forms a white ppt with Ca(OH)2 (d) To 1 cm depth of the filtrate from (b) in a test-tube (i) add a few drops of organic compound Z and warm in a water bath for 5 minutes (ii) followed by 1 cm 3 of nitric acid and 5 drops of silver nitrate. Add excess ammonia solution. Solution remains colourless. white ppt. formed, soluble in excess NH3 (e) To separate 1 cm depth of Y (i) add filtrate from (b) dropwise until in excess Pale blue ppt. formed, insoluble in excess (ii) add NaOH(aq), followed by one spatula of zinc powder and warm. (CARE!) Pale blue ppt . formed upon adding NaOH(aq). Vigourous effervescence, colourless and pungent gas evolved, which turns damp red litmus paper blue

(f) (i) carbonate / CO32‒ in residue hydroxide / OH‒ in filtrate (ii) Carboxylic acid is present. From test (c)(ii), CO2 is produced from the reaction of Z with carbonate in the residue from (b), which suggests an acidic group is present. Halogenoalkane / chlorine-containing organic compound