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 ?

Options

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

Answer

C