Question

A silver electrode is immersed in saturated $A{g}_{2}S{O}_4\left(aq\right)$. The potential difference between silver and the standard hydrogen electrode is found to be $0.711V$.
Determine the $K_{sp}$ of $A{g}_{2}S{O}_4$ at ${25}^{o}C$.
Given:
$E_{A{g}^{+}/Ag}^0=0.799V{10}^{-1.5}\approx 0.031$

• A. $K_{sp}=4.3\times {10}^{-4}$
• B. $K_{sp}=2.77\times {10}^{-4}$
• C. $K_{sp}=9.4\times {10}^{-6}$
• D. $K_{sp}=1.54\times {10}^{-5}$

Answer 1

The correct option is D $K_{sp}=1.54\times {10}^{-5}$

The cell can be represented as :
$Pt\left(s\right)|H_2(g,1atm)|H^{+}(aq,1M)||A{g}^{+}\left(salt\right)|Ag\left(s\right)$

$E_{\text{cell}}^0=E_{A{g}^{+}/Ag}^{\circ }-E_{H^{+}/H_2}^0$

$E^0=0.80-0=0.80V$
Given,
$\text{Emf of the cell}=0.711V$
The cell reaction is
$H_2+2A{g}^{+}\to 2Ag+2H^{+}$
$$Q=\frac{\left[Ag{]}^2\right[H^{+}{]}^2}{\left[H_2\right][A{g}^{+}{]}^2}=\frac{1^2\times 1^2}{1\times [A{g}^{+}{]}^2}=\frac{1}{[A{g}^{+}{]}^2}$$
Applying Nernst equation,

$E=E^0-\frac{0.0591}{2}log\frac{1}{[A{g}^{+}{]}^2}$

$0.711=0.80-\frac{0.0591}{2}log\frac{1}{[A{g}^{+}{]}^2}$

$0.089=\frac{0.0591}{2}log\frac{1}{[A{g}^{+}{]}^2}$

$3.01=log\frac{1}{[A{g}^{+}{]}^2}$

$3.01=-2log\left[A{g}^{+}\right]$

$-1.50=log\left[A{g}^{+}\right]$

$\left[A{g}^{+}\right]\approx 0.031\text{mol}{L}^{-1}$

$A{g}_{2}S{O}_4\left(aq\right)\rightleftharpoons 2A{g}^{+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)$

$\left[S{O}_4^{2-}\right]=\frac{\left[A{g}^{+}\right]}{2}$

$\left[S{O}_4^{2-}\right]\approx 0.016mol{L}^{-}$

$$K_{sp}=\left[A{g}^{+}{]}^2\right[S{O}_4^{2-}]=(0.031{)}^2\times \left(0.016\right)=1.54\times {10}^{-5}$$