Question

Find the solubility product of a saturated solution of $Ag_2CrO_4$ in water at $298 \,K$, if the emf of the cell $ Ag | Ag^+$ (Saturated $Ag_2CrO_4$ solution.) $||Ag^+ \, (0.1 M)|$ $Ag$ is $0.164\, V $ at $298 \,K$.

Answer 1

$E=0-\frac{0.0592}{1}\log \frac{[Ag^+]_\text{anode}}{[Ag^+]_\text{cathode}}$
$\Rightarrow 0.164 = - 0.0592 \log \frac{[Ag^+]+_\text{anode}}{0.10}$
$\Rightarrow [Ag^+]_\text{anode} = 1.7 \times 10^{-4}\, M $
In saturated $Ag_2CrO_4$ solution present in anode chamber :
$ Ag_2CrO_4 (s) \rightleftharpoons \, 2Ag^+ \, + CrO^{2-}_4$
$ 1.7 \times 10^{-4}\, M \, \, \, \,\, \frac{1.7}{2} \times 10^{-4}\, M $
$ K_{sp}=[Ag^+]^2[CrO^{2-}_4]$
$ = (1.7 \times 10^{-4})^2 \bigg(\frac{1.7}{2} \times 10^{-4}\bigg)$
$ = 2.45 \times 10^{-12} $