A solution 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)` values of MnS, FeS, ZnS and HgS are `10^(-15),10^(-23),10^(-20)` and `10^(-54)` respectively, which one will precipitate first?

Options

A. FeS
B. MnS
C. HgS
D. ZnS

Answer

C

Explanation

`M^(2+)+S^(2-) hArr MS [M=Mn,Fe,Zn `or `Hg]`
Metal sulphide, MS with lowest `K_(sp)` will precipitate out first. This is because its ionic product will exceed its `K_(sp)` first.