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