What `(H_(3)O^(+))` must be maintained in a saturated `H_(2)S` solution to precipitate `Pb^(2+)` , but not `Zn^(2+)` from a solution in which each ion is present at a concetration of `0.01 M ? (K_(SP)`for `H_(2)S= 1.1xx10^(-22), K_(SP)`for `ZnS=1.0xx10^(-21))`

Explanation

a. For `ZnS` not to be precipiated form a solution `Zn^(2+)` and `Pb^(2+)`
`[Zn^(2+)] [S^(2-)] lt K_(sp) of ZnS`
`(10^(-2)) [S^(2-)] lt 1.0 xx 10^(-21)`
or the maximum `[S^(2-)] = 10^(-19)` at which `ZnS` will begin to precipitate or upto this concentration, no precipitation will occur.
`H_(2)S hArr 2H^(o+) +S^(2-)`
`:. [H^(o+)]^(2) [S^(2-)] = 1.1 xx 10^(-22)`
`:. [H^(+)]^(2) [10^(-19)] = 1.1 xx 10^(-22)`
`:. [H^(o+)] = 11 xx 10^(-4) :. [H^(o+)] = 3.3 xx 10^(-2)M`
This, if `[H^(o+)] = 3.3 xx 10^(-2)` or slightly higher, the precipitation of `ZnS` will not take place and only `PbS` will precipitate.