Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\underset{\underset{1-\mathrm{s}}{1}}{\mathrm{A}{\mathrm{g}}_2\mathrm{C}\mathrm{r}{\mathrm{O}}_4}\to \underset{\underset{2\mathrm{s}}{0}}{2\mathrm{A}{\mathrm{g}}^{+}}+\underset{\underset{\mathrm{s}}{0}}{\mathrm{C}\mathrm{r}{\mathrm{O}}_4^{-2}}$
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}=\frac{(2\mathrm{s}{)}^2\mathrm{s}}{1-\mathrm{s}}$
$\mathrm{s}<<1$
As 1 >> s so 1 – s = 1
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}=4{\mathrm{s}}^3$
$=1.1\times 10^{-12}=4{\mathrm{s}}^3$
$=\mathrm{s}=6.5\times 10^{-5}$
$\mathrm{A}\mathrm{g}\mathrm{C}\mathrm{l}\to \underset{\mathrm{s}}{\mathrm{A}{\mathrm{g}}^{+}}+\underset{\mathrm{s}}{\mathrm{C}{\mathrm{l}}^{-}}$
$1.8\times 10^{-10}={\mathrm{s}}^2$
or$\mathrm{s}=1.34\times 10^{-5}$
Similarly 's' for AgBr and AgCl is found to be$7.1\times 10^{-7}$and$9\times 10^{-9}$respectively since solubility of$\mathrm{A}{\mathrm{g}}_2\mathrm{C}\mathrm{r}{\mathrm{O}}_4$is highest, so it will participate last.