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, Z

Question

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, Z

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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
`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.

4 views Answered 2023-02-05 15:29