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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 the $\mathrm{K}_{\mathrm{sp}}$ of MnS, FeS, $\mathrm{ZnS}$ and $\mathrm{HgS}$ are $10^{-15}, 10^{-23}, 10^{-20}$ and $10^{-54}$ respectively, which one will precipitate first?
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The correct answer is:
$\mathrm{HgS}$
No solution. Refer to answer key.
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