YIJC 2020 Prelim P4 Ans (1)

1 [Turn over YIJC 2020 Prelim H2 Chemistry Prelim P4 Suggested Answers 1 (b) Volume of FA 3 added to mixture /cm3 6.00 11.00 15.00 20.00 Final burette reading /cm3 27.80 20.70 15.50 8.50 Initial burette reading /cm3 0.00 0.00 0.00 0.00 Volume of FA 2 added /cm3 27.80 20.70 15.50 8.50 (c) Vmax (FA 3) = 26.25 cm3 Vmax(FA 2) = 36.00 cm3 [6] (d) gradient = 36.0 − 0.0 0.0 − 26.25 = −1.37 (e) The graph has a negative gradient because the more M2CO3 added, the lesser the amount of H2SO4 left in the mixture and hence, lesser NaOH is needed to neutralise the H2SO4. (f) (i) Since Vmax(FA 2) = 36.00 cm3, nNaOH = 0.550 × 36.0 × 10-3 = 0.0198mol nH2SO4 = 1 2 × nNaOH = 1 2 × 0.0198 = 0.0099mol cH2SO4 = 0.0099 25.0 × 10-3 = 0.396 mol dm-3 (ii) nM2CO3 = nH2SO4 = 0.0099 mol Since Vmax(FA 3) = 26.25 cm3, cM2CO3 = 0.0099 26.25 × 10-3 = 0.377 mol dm-3

2 (iii) Mr of M2CO3 = 39.75 0.377= 105.4 Ar of M = 105.4-12.0-(16.0×3) 2 = 22.7 (g) Some NaOH will react with CO 2 in the air. Amount of NaOH left in FA 2 solution will decrease /[NaOH] decreases, resulting in a larger value of Vmax (FA 2) required to neutralise the H2SO4 solution. (h) % error for mixture 1 = 0.05 × 2 27.80 × 100% = 0.360% % error for mixture 2 = 0.05 × 2 8.50 × 100% = 1.18% The percentage error is smaller as the titre volume for mixture 1 is larger than that of mixture 2. (i) (i) Experiment Volume of FA 5 /cm3 Volume of FA 6 /cm3 Initial temperature of FA 5 /oC Maximum temperature /oC Maximum change in temperature/oC 1 20.0 40.0 30.4 37.4 7.0 2 40.0 20.0 30.8 38.1 7.3 (ii) Acid X is dibasic. Amount of water formed in experiment 2 is same as that in experiment 1. Heat released for experiment 2 is same as that of experiment 1. For the same volume of solution, change in temperature will be the same. 2 (a) Experiment Volume of FA 7 /cm3 Volume of FA 8 /cm3 Volume of FA 10/cm3 volume of deionised water/cm3 t/s 1/t/s-1 1 5.0 5.0 10.0 10.0 69 0.0145 2 10.0 5.0 10.0 5.0 35 0.0286 3 5.0 10.0 10.0 5.0 34 0.0294 4 5.0 5.0 20.0 0.0 18 0.0556 (b) Total volume should be constant so that the concentration of reactants is directly proportional to the volume used. (c) • [BrO3–] Comparing experiments 1 and 2, When [BrO3–] doubles, (keeping [Br–] and [H+] constant), relative rate doubles. Hence, order of reaction with respect to [BrO3–] is one. • [Br–] Comparing experiments 1 and 3, When [Br–] doubles, (keeping [BrO3–] and [H+] constant), relative rate doubles. Hence, order of reaction with respect to [Br–] is one. • [H+] Comparing experiments 1 and 4,

3 When [H+] doubles, (keeping [BrO3–] and [Br–] constant), relative rate quadruples. Hence, order of reaction with respect to [H+] is two. (d) Rate = k [BrO3–][Br–][H+]2 (e) [phenol] = 15 ×0.0301000 50 1000 = 9.00 × 10-3 mol dm-3 (f) (decrease) change in [BrO3–] = 1 3 × [phenol] = 3.00 × 10-3 mol dm-3 Rate = 3.00 × 10-3/ time in expt 1 = 3.00 ×10-3 / 69 = 4.35 × 10-5 mol dm-3 s-1 (g) Rate = k [BrO3–][Br–][H+]2 4.35 × 10-5 = k( 5 50 × 0.350)( 5 50 × 0.650)( 10 50 × 0.250 × 2)2 k = 1.91 mol-3 dm9 s-1 (h) Any change in the rate of reaction is not due to change in concentration of FA 9. OR Volume of FA 9 should be kept constant as we are measuring the time taken for a fixed amount of Br 2 to be used up for the complete monobromination of FA 9 (and relating this time to the relative rate) (i) The bromo substituent is electron withdrawing / ring deactivating . This decreases the electron density in the ring and makes it less susceptible to electrophilic attacks. 3 (a) Ksp = [Ca2+] [OH–]2 (b) 1. Using a 100.0 cm3 measuring cylinder, measure 100.0 cm3 of 0.100 moldm-3 NaOH (aq) into a 250 cm3 beaker. 2. Using a spatula, add solid Ca(OH)2 to the beaker containing NaOH(aq) until no more solid is seen to dissolve. (to prepare a saturated solution) 3. Stir well with a glass rod (for three minutes). 4. Allow the excess solid to settle and filter the mixture, using dry filter paper and dry filter funnel, into a dry 250 cm3 conical flask. 5. Keep the filtrate. 6. Fill the burette with 0.100 moldm-3 HCl. 7. Pipette 25.0 cm3 of the filtrate into a clean 250cm3 conical flask. 8. Add 2 – 3 drops of phenolphthalein / methyl orange / thymol blue indicator.

4 9. Titrate with HC l until the solution changes from pink to colourless/ yellow to orange / blue to colourless. 10. Repeat the titration until two consistent titres of within +/- 0.10 cm3 are obtained. (c) (i) Amount of OH– = (26.05/1000) × 0.100 = 0.002605 mol [OH–] = 0.002605 / 0.025 = 0.104 moldm-3 (ii) [OH–] from Ca(OH)2 = [OH–] total – [OH–] NaOH = 0.1042 – 0.100 = 4.20 × 10-3 mol dm-3 (iii) [Ca2+] = 0.0042/2 = 2.10 × 10-3 mol dm-3 (iv) Ksp = [Ca2+] [OH–]2 = 2.10 × 10-3 × (0.1042)2 = 2.28 x 10-5 mol3 dm-9 (d) As NaOH is a strong base, it will fully dissociate to give Na+ and OH‒. Hence, this increases the overall concentration of OH‒ in the solution which is a common ion. As a result, according to LCP, the position of equilibrium Ca(OH) 2(s) ⇌ Ca2+(aq) + 2OH –(aq) will shift to the left, resulting in a decrease in solubility of calcium hydroxide. (e) pH VHCl/cm3 13