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A solution which is 10^(-3) M each in Mn...

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

MgS

C

HgS

D

ZnS

Text Solution

Verified by Experts

Hgs having the lowest `K_(sp)` among the lot will precipitate first.
Hence, (C) is the correct answer.
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