Chemistry 9701 · AS & A Level · Equilibria

Equilibria — practice question

Silver carbonate, $\text{Ag}_2\text{CO}_3$, dissolves only slightly in water. Its solubility product, $K_{sp}$, is $6.3 \times 10^{-12}$ at $25^\circ\text{C}$.
(a(i))[2]

State an expression for the solubility product, $K_{sp}$, of $\text{Ag}_2\text{CO}_3$, and give its units.

(a(ii))[1]

For a saturated solution of $\text{Ag}_2\text{CO}_3$ at $25^\circ\text{C}$, calculate the equilibrium concentration of $\text{Ag}^$.

(a(iii))[1]

Solid $\text{Ag}_2\text{CO}_3$ is stirred at $25^\circ\text{C}$ with $0.050\ \text{mol dm}^{-3}\ \text{AgNO}_3$ until no more $\text{Ag}_2\text{CO}_3$ dissolves. Calculate the concentration of carbonate ions, $[\text{CO}_3^{2-}]$, in the resulting solution.

(a(iv))[2]

An electrochemical cell is assembled to measure the electrode potential, $E$, of the $\text{Ag}^+/\text{Ag}$ half-cell, using saturated $\text{Ag}_2\text{CO}_3(\text{aq})$ together with a standard hydrogen electrode. Use the Data Booklet, your answer to part (a)(ii), and the Nernst equation to calculate the electrode potential, $E$, for this $\text{Ag}^+/\text{Ag}$ half-cell.

(b(i))[1]

Calculate the standard entropy change of solution, $\Delta S^0b0$, for the process $\text{AgCl}(s) \rightarrow \text{Ag}^+(aq) + \text{Cl}^-(aq)$.

(b(ii))[3]

Explain, with the aid of a calculation, why $\text{AgCl}$ is insoluble in water at $25^\circ\text{C}$. You should use data from this question and your answer to part (b)(i).

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: The expression is $K_{sp} = [Ag^+]^2[CO_3^{2-}]$.

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