A-LevelChemistryEquilibria9701/41 · Oct/Nov 2019 · Q59 Marks

Silver sulfide, Ag₂S, is very insoluble in water.

(a)(i)Write an expression for the solubility product, Ksp, of Ag₂S(s). Ksp = .......................................................................... [1][1]
(a)(ii)The solubility of Ag₂S(s) in water at 298 K is 1.16 × 10⁻¹⁷ mol dm⁻³. Calculate the numerical value of the solubility product, Ksp, of Ag₂S(s) at 298 K. Ksp = .......................................................................... [2][2]
(a)(iii)Calculate the minimum volume of water needed to dissolve 1.00g of Ag₂S(s) under standard conditions. volume = .......................................................... dm³ [2][2]
(b)(i)Calculate the pH of 0.20 mol dm⁻³ HOBr(aq). pH = .......................................................................... [2][2]
(b)(ii)5.0 cm³ of 0.20 mol dm⁻³ potassium hydroxide, KOH, are added to 20.0 cm³ of 0.20 mol dm⁻³ HOBr(aq). Calculate the pH of the buffer solution produced. pH = .......................................................................... [2][2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[1] Ksp = [Ag+]2[S2-]
[2] • [S2-] = 1.16 × 10-17 • [Ag+] = 2.32 × 10-17 • Ksp = 6.2(4) × 10–51 minimum 2 sig. fig. Award 1 mark for two points, award 2 marks for three points
[2] M1: moles Ag2S = 1 / 247.9 = 0.00403 moles [1] 2sf min M2: 1.16 × 10-17 = 0.0040 / V so V = 3.5 × 1014 (dm³) [1] 2sf min ecf on M1
[2] M1: [H+] = √2.0 × 10¯º × 0.20 [H+] = 2.0 × 10-5 (1.9976 × 10-5) M2: pH = 4.7 (4.699) minimum 2 sig. fig. min
[2] M1: Both equilibria correctly stated moles KOH = 0.005 × 0.2 = 1 x 10-3 moles HOBr(initial) = 0.020 × 0.2 = 4 × 10-3 moles HOBr(eqm) = 4 × 10–3 – 1 × 10-3 = 3 x 10-3 moles BrO-(eqm) = 1 × 10–3 M2: ratio [OBr]/[HOBr] = 1/3 [H+] = 3 x 2.0 × 10–9 = 6 × 10–9 pH = 8.2(2)

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