2021 RI H2 Chem Prelims P4 Answers

-1- (b)(ii) Average titre volume = 22.75+22.75 2 = 22.75 cm3 (2 d.p.) (c)(i) Amount of S2O32− ions = 0.100  22.75 1000 = 2.275  10−3 = 2.28  103 mol (3 s.f.) (c)(ii) Amount of I2 reacted = ( 1 2) (2.275  103) = 1.138 x 10−3 mol Amount of IO3– ions in 25.0 cm3 of FA 1 = ( 1 3)(1.138  10−3) = 3.793  10−4 = 3.79  10−4 mol (3 s.f.) (c)(iii) Amount of IO3– ions in saturated solution prepared in (a) =( 100.0 25.0 )(3.793  10−4) = 1.517  10−3 = 1.52  10−3 mol (3 s.f.) (c)(iv) Amount of IO3– from KIO3 = 0.0100  50.0 1000 = 5.00  104 mol Amount of IO3– ions from the Ca(IO3)2 that has dissolved = 1.517  10−3 – 5.00  104 = 1.017  10−3 = 1.02  10−3 mol (3 s.f.) (c)(v) Amount of Ca2+ ions in saturated solution prepared in (a) = ( 1 2) (1.017  10−3) = 5.085  10−4 = 5.09  10−4 mol (3 s.f.) (c)(vi) Ksp of Ca(IO3)2 = [Ca2+] [IO3−]2 units: mol3 dm−9 (c)(vii) Ksp of Ca(IO3)2 = (5.085 x 10−4 0.100 ) (1.517 x 10−3 0.100 ) 2 = 1.17 x 10−6 mol3 dm−9 (3 s.f.) (d) The concentration of IO3– in FA 1 is lowered . This forms less er I2 for reaction with S2O32−. Therefore, the titre volume would be lower. (e)(i) Total [ions] in solvent prepared for Experiment 1 = 2 x 0.0100 = 0.0200 mol dm-3 Total amount ions in solvent prepared for Experiment 2 = 2(0.0100  25.0 1000) + 2 (0.0100  75.0 1000) = 0.002 mol Total [ions] in solvent prepared for Experiment 2 = 0.002 0.100 = 0.0200 mol dm-3 2021 Y6 H2 Chemistry Prelim Exam Paper 4 – Suggested Solutions 1(b)(i) Titration number 1 2 Final burette reading / cm3 22.75 22.75 Initial burette reading / cm3 0.00 0.00 Volume of FA 2 used / cm3 22.75 22.75 Values used  

-2- The total concentration of ions in both solvents are the same. (e)(ii) As the volume of KIO3(aq) used increases, the concentration of IO3 increases and further suppresses the dissociation of calcium iodate. The solubility of Ca( IO3)2 decreases, hence the amount of IO3– ions obtained from Ca(IO3)2 that has dissolved will decrease. (e)(iii) Do not agree. volume of KIO3(aq) used 2(a) The dissolving of KIO3 is endothermic, as an increase in temperature shifts the equilibrium position to the right , favouring the endothermic dissolving process to remove some of the extra heat supplied. (b) Solubility = 100 V × mass of KIO3 = 100 4.0 × 1.00 = 25.0 g per 100 g of water (c) Solubility = 100 V × mass of KIO3 6.25 = 100 V × 1.00  V = 16.0 cm3 Hence the volume of water that will just dissolve 1.00 g KIO3 at 10 C is 16.0 cm3. (d) Procedure: 1. Weigh an empty boiling tube using a weighing balance and record its mass. 2. Transfer the contents of the weighing bottle with KIO3 to the boiling tube. Reweigh the boiling tube and its contents using a weighing balance and record the mass. 3. Use a 10 cm 3 measuring cylinder (or burette) to transfer 4.0 cm 3 of deionised water to the weighed boiling tube with KIO3. 4. Prepare a hot water bath by filling a 250 cm3 beaker half full of water and heat it over the Bunsen burner. 5. Warm the boiling tube carefully in the water bath, while stirring the contents with a thermometer, until all the solid has dissolved. 6. Clamp the tube on a retort stand. Let the boiling tube cool and continue to stir gently with the thermometer. 7. Note and record the temperature at which crystals are first formed in the solution. 8. Add a further 2.0 cm 3 of deionised water to the boiling tube using the 10 cm 3 measuring cylinder/burette. 9. Warm the boiling tube in the water bath to re-dissolve the solid, and repeat steps 6 and 7. 10. Repeat steps 8, 5, 6 and 7 until five (if 3 cm 3 portions are added) OR seven readings (if 2 cm3 portions are added) in total are obtained (with the lowest temperature at about 10 C). Ksp of calcium iodate(V)

-3- 11. If crystals do not appear at temperatures close to room temperature, add ice cubes to the water bath to further lower the temperature. 12. Calculate the corresponding solubilities at the different temperatures. (e) Weighing by difference for boiling tube: Mass of boiling tube + KIO3 / g m1 Mass of empty boiling tube / g m2 Mass of KIO3 used / g m1  m2 OR Weighing by difference for weighing bottle: Mass of weighing bottle + KIO3 / g m1 Mass of emptied weighing bottle / g m2 Mass of KIO3 used / g m1  m2 OR Use of TARE for boiling tube: Mass of KIO3 used = m g (TARE) Table for volume of water and temperature: Total volume of water, V / cm3 Temperature at first appearance of crystals/ oC 4 T 6 8 10 12 14 16 Solubility s at temperature T = 100 V × (m1 − m2) or 100 V × m 3(a)(i) Mass of bottle and FA 5 / g 6.095 Mass of emptied bottle / g 5.222 Mass of FA 5 / g 0.873 t / min T / °C 0.0 28.4 1.0 28.4 2.0 28.4 3.5 43.4 4.0 51.4 5.0 51.3 6.0 50.9 7.0 50.3 8.0 49.6 9.0 49.2

-4-

-5- (a)(ii) Plot a graph of T / °C against t / min 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 T / °C t / min

-6- (a)(iii) Tmin = 28.4 C Tmax = 52.5 C T = +24.1 C (a)(iv) q = (50.0) (4.18) (24.1) = 5036.9 J = 5040 J or 5.04 kJ (a)(v) Using 100 cm3 of FA 4 would lead to the value of T being halved. However, since m is doubled, this would lead to the same value of q. (b)(i) M cannot be calcium because calcium sulfate is insoluble and no ppt was observed in the final mixture. (b)(ii) M is Mg. (c) Mass of MgO in FA 5 = (0.87)(0.873) = 0.7595 g Mass of Mg in FA 5 = (0.13)(0.873) = 0.1135 g q = qMg + qMgO 5.0369 = (– 466.8) − 0.1135 24.3 + H1 (− 0.7595 40.3 ) 5.0369 = 2.180 – (0.01871) H1 H1 = −153 kJ mol−1 4(a) Table 4.1 Test observations 1 Test the solution of FA 6 with Universal Indicator Paper. pH = 2 or 3 dark orange / orange 2 Add 1 cm depth of FA 6 to a boiling tube. Add aqueous sodium hydroxide until no further change occurs, then warm the boiling tube gently. No ppt / No NH3 evolved / no observable change Allow the mixture to cool slightly. Add 1 piece of alumin ium foil, then warm the boiling tube gently. No NH3 evolved / no observable change. 3 Add 1 cm depth of FA 6 to a test-tube. Add 1 cm depth of aqueous barium nitrate. Add dilute nitric acid dropwise until no further change occurs. White ppt formed. White ppt insoluble in dilute nitric acid.

-7- (b)(i) Table 4.2 identity evidence anion P SO42– In test 3, a white ppt of BaSO4 was formed which was insoluble in nitric acid. cation Q H+ In test 1, the pH of the solution is acidic. In test 2, when FA 6 was heated with NaOH, there was no ppt/no observable change. (b)(ii) S2O82– (c)(i) A red-brown ppt is formed. There is effervescence of CO2 which gave a white ppt in limewater. (c)(ii) Fe3+ Red-brown ppt of Fe(OH)3 formed. There is effervescence of CO2 which gave a white ppt in limewater. (d)(i) Solution in test tube A turned brown faster than the solution in test tube B. (d)(ii) FA 6 is an oxidising agent as it oxidises iodide in potassium iodide to the brown iodine observed. (d)(iii) Table 4.3 test-tube A test-tube B Brown solution decolourises Red-brown ppt formed. Brown solution decolourises (d)(iv) FA 7 is a homogeneous catalyst. From (i), iodide was oxidised to brown iodine more quickly in test -tube A than B, hence FA 7 sped up the reaction rate in test-tube A. From (iii), the formation of red -brow