A solution containing both `Zn^(2+)` and `Mn^(2+)` ions at a concentration of `0.01M` is saturated with `H_(2)S`. What is pH at which MnS will form a ppt ? Under these conditions what will be the concentration of `Zn^(2+)` ions remaining in the solution ? Given `K_(sp)` of `ZnS is 10^(-22)` and `K_(sp)` of MnS is `5.6 xx 10^(-16), K_(1) xx K_(2)` of `H_(2)S = 1.10 xx 10^(-21)`.

Explanation

The minimum `[S^(2-)]` for the start of the precipitaitons is that which satisfies the `K_(sp)` of `MnS` is in which `[Mn^(2+)] = 0.01M`
`[S^(2-)] = K_(sp) "of"(MnS)/([Mn^(2+)]) =(5.6 xx 10^(-16))/(0.01) = 5.6 xx 10^(-14)`
The `[H^(o+)]` of the soln. having the above `[S^(2-)]` can be calculated form the `H_(s)S` dissociation eq. expression.
`([H^(o+)]^(2)[S^(2-)])/([H_(2)S]) = ([H^(o+)]^(2)xx5.6xx10^(-14))/(0.01) =1.1 xx 10^(-21)`
`:. [H^(o+)] = 4.43 xx 10^(-5)` and `pH = 4.35`
If the `[H^(o+)]` is greater than `4.34 xx 10^(-5)M` than the `[S^(2-)]` will be less than `5.6 xx 10^(-14)M` and `MnS` will no longer ppt. from the soln. `:.` The conce of `Zn^(2+)` ion remaining in the soln. can be calculated form the `K_(sp)` of `ZnS`.
`:. [Zn^(2+)] = (K_(sp) "of" ZnS)/([S^(2-)]) = (1.0 xx 10^(-22))/(5.6 xx 10^(-14)) = 1.76 xx 10^(-9)M`
Thus, by peroply adjusting the `[H^(o+)]` in the soln. it is possible to precipitate effectively all of the `Zn` from the solution without precipitation any `Mn^(2+)` ion.