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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

1 Answer

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Best answer
Correct Answer - C
HgS having the lowest `K_(SP)` among the lot will precipitate first.

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