Question

What `[H^(+)]` must be maintained in a saturated `H_(2)S(0.1)M` to precipitate `CdS` but not `ZnS` if `[Cd^(2+)]=[Zn^(2+)]=0.1` initially?

Answer 1

In order to prevent precipitaion of `ZnS`,
`[Zn^(2+)][S^(2-)] ltK_(sp)(ZnS)=1xx10^(-21)`
(Ionic product)
or `(0.1.[S^(2-)]lt 1xx10^(-21)`
or `[S^(2-)]lt 1xx10^(-20)M`
This is the maximum value of `[s^(2-)]` before `ZnS` will precipitate .Let `[H^(+)]` to maintain this `[S^(2-)]` be `x`.
Thus for `H_(2)S hArr 2H^(+)+S^(2-)`
`K_(a)=([H^(+)]^(2)[S^(2-)])/([H_(2)S])=(x^(2)(1xx10^(-20))a)/(0.1)=1.1xx10^(-21)`
or `x=[H^(+)]=0.105M`
No `ZnS` will precipitate of concentration of `H^(+)` greater than `0.105M`.