Chemistry - Ionic Equilibrium Question with Solution | TestHub

For the precipitation of MS form a solution of$\mathrm{M}\mathrm{B}{\mathrm{r}}_2$by passing${\mathrm{H}}_2\mathrm{S}$
${\text{K}}_{\text{sp}\left(\text{MS}\right)}=\left[{\text{M}}^{2+}\right]\left[{\text{S}}^{2-}\right]$
or$6\times {10}^{-21}=\left[0.05\right]\left[{\mathrm{S}}^{2-}\right]$
or$\left[{\mathrm{S}}^{2-}\right]=1.2\times {10}^{-19}\mathrm{M}$
i.e. precipitation of MS will start when${\mathrm{H}}_2\mathrm{S}$provides
$1.2\times {10}^{-\mathrm{19 }}\mathrm{M }{\text{ion of S}}^{2-}$
For${\text{H}}_2\text{S}\rightleftharpoons 2{\text{H}}^{+}+{\text{S}}^{2-}$
$\mathrm{K}={\mathrm{K}}_1\times {\mathrm{K}}_2=\frac{{\left[{\mathrm{H}}^{+}\right]}^2\left[{\mathrm{S}}^{2-}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]}$
or${10}^{-7}\times 1.3\times {10}^{-13}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2[1.2\times {10}^{-19}]}{0.1}$
or$\left[{\mathrm{H}}^{+}\right]=1.04\times {10}^{-1}$
$\therefore \text{pH=0.9826}$