The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is `1.0 xx 10^(-19)` M. If `10 mL` of this is added to `5 mL` of `0.04 M` solution of the following: `FeSO_(4), MnCl_(2), ZnCl_(2)` and `CdCl_(2)`. in which of these solutions precipitation will take place?

Explanation

Precipitation will take place in the solution for which ionic product is greater than solubility product. As `10 mL` of solutio containing `S^(2-)` ion is mixed with `5 mL` of metal salt solution, after mixing
`[For[S^(2-)]`:
`M_(1)V_(1)=M_(2)V_(2)`
`10^(-19)xx10=M_(2)xx15, M_(2)=[S^(2-)]=6.67xx10^(-20) M`
For
`[Fe^(2+)]=[Mn^(2+)]=[Zn^(2+)]=[Cu^(2+)]:`
`M_(1)V_(1)=M_(2)V_(2)`
`0.04xx5=M_(2)xx15 rArr M_(2)=1.33xx10^(-2) M`
i. `Q_(sp)=[Fe^(2+)][S^(2-)]=1.33xx10^(-2)xx6.67xx10^(-20)`
`=8.77xx10^(27)`
`[Q_(sp) lt K_(sp)`, no precipitation]
`:.` Ionic product each of these will be
`=[M^(2+)][S^(2-)]=8.77xx10^(-27)`
As this greacter then the `K_(sp)` of `ZnS`, and `CdS`, therefore `ZnS` and `CdS` solution will be precipitated and `FeS` and `MnS` will not be precipitated.