Chemical and Ionic Equilibrium 2 Question 17
17. A solution which is $10^{-3} \mathrm{M}$ each in $\mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Zn}^{2+}$ and $\mathrm{Hg}^{2+}$ is treated with $10^{-16} \mathrm{M}$ sulphide ion. If $K_{\mathrm{sp}}$ of $\mathrm{MnS}, \mathrm{FeS}, \mathrm{ZnS}$ and $\mathrm{HgS}$ are $10^{-15}, 10^{-23}, 10^{-20}$ and $10^{-54}$ respectively, which one will precipitate first?
(2003, 1M)
(a) $\mathrm{FeS}$
(b) $\mathrm{MgS}$
(c) $\mathrm{HgS}$
(d) $\mathrm{ZnS}$
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Solution:
- Minimum $\mathrm{S}^{2-}$ concentration would be required for precipitation of least soluble $\mathrm{HgS}$.
For $\mathrm{HgS}, \mathrm{S}^{2-}$ required for precipitation is
$$ \left[\mathrm{S}^{2-}\right]=\frac{K_{\mathrm{sp}}}{\left[\mathrm{Hg}^{2+}\right]}=\frac{10^{-54}}{10^{-3}}=10^{-51} \mathrm{M} $$
