A-LevelChemistryEquilibria9701/43 · May/June 2020 · Q710 Marks

Silver carbonate, Ag2CO3, is sparingly soluble in water. The numerical value of the solubility product, Ksp, for silver carbonate is 6.3 × 10–12 at 25°C.

(a)(i)Write an expression for the solubility product, Ksp, of Ag2CO3, and state its units. Ksp = .............................................. units = ..............................................[2]
(a)(ii)Calculate the equilibrium concentration of Ag+ in a saturated solution of Ag2CO3 at 25°C. [Ag+] = .............................................. mol dm-³[1]
(a)(iii)Solid Ag2CO3 is stirred at 25 °C with 0.050 moldm¯³ AgNO3 until no more Ag2CO3 dissolves. Calculate the concentration of carbonate ions, [CO32-], in this solution. [CO32-] = .............................................. mol dm-³[1]
(a)(iv)An electrochemical cell is set up to measure the electrode potential, E, for the Ag+/Ag half-cell using the saturated Ag2CO3(aq) with a standard hydrogen electrode. Use the Data Booklet, your answer to (a)(ii), and the Nernst equation to calculate the electrode potential, E, for this Ag+/Ag half-cell. E for Ag+/Ag half-cell = .............................................. V[2]
(b)(i)Silver chloride, AgCl, is sparingly soluble in water. The equation for the enthalpy change of solution is shown. AgCl(s) → Ag+(aq) + Cl-(aq) ∆H°sol = +65.5 kJ mol-1 Standard entropies are shown in the table. [Table with species and S°/J K-1 mol-1 values for AgCl(s), Ag+(aq), Cl-(aq)] Calculate the standard entropy change of solution, ∆S°. ∆S° = .............................................. J K-1 mol-1[1]
(b)(ii)Explain, with the aid of a calculation, why AgCl is insoluble in water at 25 °C. You should use data from this question and your answer to (b)(i).[3]

Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

[2] (a)(i) M1 Ksp = [Ag+]2[CO32-]
[2] (a)(i) M2 units = mol³ dm-9
[1] (a)(ii) x = ³√6.3 x 10-12/4 = 1.16 x 10-4 (mol dm-³)
[1] (a)(ii) [Ag+] = 1.16 x 10-4 x 2 = 2.33 x 10-4 (mol dm-³) min 2sf
[1] (a)(iii) 6.3 x 10-12 = [0.05]2[CO32-]
[1] (a)(iii) [CO32-] = 2.52 x 10-9 (mol dm-³) min 2sf
[2] (a)(iv) M1 E=E° + 0.059log[Ag+]
[2] (a)(iv) M2 E=0.80 + 0.059log(1.2 x 10-4) = 0.57 V ecf from (a)(ii) min 2sf
[1] (b)(i) ∆S° = 72.7 + 56.5 -96.2 = +33.0 J K-1 mol-1
[3] (b)(ii) M1 ∆G = ∆H° - T∆S°
[3] (b)(ii) M2 ∆G = (65.5)- (298x0.033) = +55.7 kJ mol-1 min 3sf
[3] (b)(ii) M3 ∆G = positive so not feasible/spontaneous