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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+)` it treated with `10^(-16)M` sulphide ion. If the `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

Text Solution

Verified by Experts

The correct Answer is:
C
`[S^(2-)]` needed for precipitation of
`FeSimplies(10^(-23))/(10^(-3))=10^(-20)M`
`Mnsimplies(10^(-15))/(10^(-3))=10^(-12)M`
`ZnSimplies(10^(-20))/(10^(-3))=10^(-17)M`
`HgSimplies(10^(-54))/(10^(-3))=10^(-51)M`
Thus, minimum `[S^(2-)]` is for HgS and so it will precipitate first.
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