2019 HCI Prelim H2 Chem P4 ANS

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 Paper 4 1 (a) (i)  Tabulates volumes and temperature data in (a)(i) Table should contain correct headers and units. Data headers to include volume of FA 1, volume of FA 2, TFA1, Tmax and T. [1] Volume of FA 1 / cm3 Volume of FA 2 / cm3 TFA 1 / °C Tmax / °C T / °C 10.0 40.0 32.0 36.0 4.0 20.0 30.0 32.0 39.5 7.5 25.0 25.0 32.0 41.5 9.5 30.0 20.0 32.0 41.0 9.0 35.0 15.0 32.0 38.5 6.5 40.0 10.0 32.0 36.0 4.0  Complete 6 sets of volume/temperature readings in (a)(i) Required volumes: Volume of FA 1 = 10.0 cm3, 20.0 cm3, 25.0 cm3, 30.0 cm3, 35.0 cm3 and 40.0 cm3 and use appropriate volume of FA 2 such that the total volume of reacting mixture in each set of data is 50 cm3 [1]  Records all temperature data in (a)(i) to 0.5°C, all volumes for FA 1 and FA 2 in (a)(i) to 1 d.p. [1]  Correctly calculates all T values to 1 d.p. in (a)(i) [1] (a) (ii)  Axes correct way round and correct labels and units and scale [1] Note: Scale chosen must allow for the lines to be extrapolated to cross each other. The plotted points should occupy at least half the grid in both directions.  All points are correctly plotted to within ± ½ small square. [1]  All drawn graph lines are straight best-fit lines and are extrapolated to cross each other. and there are at least three points on each side of the graphically determined Tmax [1] HWA CHONG INSTITUTION 2019 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS

(a) (iii) Tmax and Vmax are read correctly to ± ½ small square from graph From the sample graph, Tmax = 10.4 °C Vmax = V(H2SO4) = 27.25 cm3 [1] (b) (i) H2SO4 + 2 NaOH → Na2SO4 + 2 H2O V(NaOH) = 50.0 – 27.25 = 22.75 cm3 n(NaOH) = 1.50 × (22.75/1000) = 0.03413 mol n(H2SO4) = 0.03413 / 2 = 0.01706 mol [H2SO4] = 0.01706 / (27.25/1000) = 0.626 mol dm−3 [1] (ii) Heat change = msolutioncTmax = 50.0 × 4.18 × 10.4 = 2174 J [1] (c) n(H2O) = n(NaOH) = 0.03413 mol Hneut = − 2174 / 0.03413 = − 6.37 × 104 J mol−1 = − 63.7 kJ mol−1 (Sign must be negative) [1] (d) Hneut would be less exothermic as malonic acid is a weak acid. Energy is absorbed to ionise the un-ionised weak acid. [1] (e) Suggested sources of errors and its appropriate improvements: [1] heat loss to the surrounding and use a cup lid to minimise heat exchange with the surrounding air VFA 1 and VFA 2 is to 1 d.p. as measured using less precise measuring cylinders and can be measured using more precise burettes to give 2 d.p. initial temperature of FA 2 was not accounted for and weighted initial temperature should be calculated where Tweighted initial = (Volume of FA 1 x TFA1) + (Volume of FA 2 x TFA2) Volume of FA 1 + Volume of FA 2 heat capacity of the calorimeter (Styrofoam cup) was not accounted for and heat absorbed by the styrofoam cup can be included in the calculation of heat change 2 (a) Mass of weighing bottle and FA 3 / g Mass of weighing bottle and residual FA 3 / g Mass of FA 3 used / g Tables have correct headers and units (included in the header or with each entry in the table) [1]

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 (b) Effervescence observed. [0.5] and Yellow / greenish yellow / yellow green solution turned colourless / very pale green. [0.5] (c) (i) Final burette reading / cm3 Initial burette reading / cm3 Volume of FA 4 used / cm3  Tables have correct headers and units (included in the header or with each entry in the table) [1] Note: Mark is lost if any final and initial burette readings are inverted or 50 is used as the initial burette reading.  All mass reading in (a) are recorded to the nearest 0.01 g and burette readings & volume used for all accurate titres in the titration table are recorded to the nearest 0.05 cm3. [1]  At least two uncorrected titres for end-point within ±0.10 cm3. [1] (c) (ii)  Student obtains average titre, to 2 d.p., from any experiments with end- point titre values within 0.10 cm3 [1] Mark is lost if there are arithmetic errors in the table. Mark is lost if the titres used are not identified either in the table (by, for example, a tick) or in a calculation.  Accuracy Supervisor’s VFA 4 / mFA 3 = 1.530 Difference between student’s and supervisor’s VFA 4 / mFA 3 If difference is  0.045 cm3 g1 [2] If difference is > 0.045 but  0.075 cm3 g1 [1] For a difference > 0.075 [0] (d) (i) If VFA 4 = 10.10 cm3 (Fe2+) = 10.10  103  0.020  5 = 1.01  103 mol (3 s.f) [1] (d) (ii) (Fe3+) = 1.01  103  250 / 10.0 = 0.0253 mol (3 s.f) [1] (d) (iii) mFA 3 = 6.60 g Mole ratio of Fe2(SO4)3.nH2O : Fe3+ = 1 : 2 Number of moles of Fe2(SO4)3.nH2O = 0.0253  2 = 0.01265 mol Mr of Fe2(SO4)3.nH2O = 6.60 / 0.01265 = 521.7 [1] n = 521.7  [2(55.8) + 3(32.1) + 12(16.0)]  18.0 = 7 (nearest whole number) [1]

 Shows working in 1(b)(i), 1(b)(ii), 1(c), 2(d)(i), 2(d)(ii) and 2(d)(iii). All calculations must be relevant although they may not be complete or correct. Any calculation not attempted loses this mark. [1]  Shows appropriate significant figures (3 or 4 sf) in all final answers (in the blank) in 1(b)(i), 1(b)(ii), 1(c), 2(d)(i) and 2(d)(ii). For 2(d)(iii), Mr can be given to 1 d.p. or 3 s.f. but n should be given as a whole number. Any calculation not attempted loses this mark. [1]  Shows appropriate units in all final answers (in the blank) in 1(a)(iii) (C, cm3), 1(b)(i) (mol dm3), 1(b)(ii) (J or kJ), 1(c) (J mol1 or kJ mol1), 2(c)(ii) (cm3), 2(d)(i) (mol), 2(d)(ii) (mol). Units should not be given for Mr or n in 2(d)(iii). Any calculation not attempted loses this mark. [1] (e) Effervescence was observed as zinc reacted with acid to form H2 gas. [0.5] Yellow solution turned colourless / very pale green as Fe3+ was reduced to Fe2+. [0.5] (f) Zinc metal that is not removed will continue to reduce Fe 3+ formed during the titration to Fe2+, resulting in a higher than expected titre. [1] 3 (a) Kc = [FeSCN2+] [Fe3+][SCN–] [1] (b) no. of moles of FeSCN2+ formed  no. of moles of SCN– added = 2.00 x 10–3 x (5/1000) = 1.00 x 10–5 mol [FeSCN2+] = (1.00 x 10–5)/(10/1000) = 1.00 x 10–3 mol dm–3 [1] Large excess of Fe3+ is used to drive the equilibrium almost completely to the right. [1] Hence, the amount of FeSCN 2+ produced will be essentially equal to the amount of SCN– added. (c) Procedure (sample answer) 1. Using separate burettes, transfer 5.00 cm3 of Fe(NO3)3 and 5.00 cm3 of KSCN into a boiling tube. Stopper and shake to ensure a homogeneous solution. 2. Rinse and fill a cuvette with 3 cm3 of the mixture. 3. Place the cuvette in a thermostatically controlled water bath maintained at 50 °C for about 5 min. 4. Measure and record the temperature of the solution using a thermometer.

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 4 5. Remove the cuvette from the water bath, and immediately measure and record the absorbance (at wavelength of blue light 447 nm) using a spectrophotometer. 6. Repeat steps 3 – 5 at 40 °C, 30 °C, 20 °C and 10 °C. or Repeat steps 2 – 5 (total volume must then be at least 5×3=15 cm3) at 40 °C, 30 °C, 20 °C and 10 °C. M1 – Logical sequence [1] (minus ½ mark for each missing point)  Mixing known volumes of Fe(NO3)3 and KSCN, amount of Fe3+ not more than 10 times of SCN–, total volume > 3 cm3  Measure (and record) the temperature of the mixture  Measure (and record) the absorbance of the mixture M2 – Apparatus [1] (½ mark for each point)  separate burettes for Fe(NO3)3 and KSCN ( or other precise apparatus e.g. micropipette or pipette)  thermostatically controlled water bath M3 – Essential details [1] (½ mark for each point)  Shake boiling tube / test-tube/ swirl conical flask (with