$A$ solution which is $10^{-3} \ M$ each in $Mn^{2+}, Fe^{2+}, Zn^{2+},$ and $Hg^{2+}$ is treated with $10^{-16} \ M$ sulphide ion. If $K_{sp}$ of $MnS, FeS, ZnS,$ and $HgS$ are $10^{-15}, 10^{-23}, 10^{-20},$ and $10^{-54}$ respectively,which one will precipitate first?
- A$FeS$
- B$MnS$
- C$HgS$
- D$ZnS$
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