1.

What `[H_(3)O^(+)]` must be maintained in a saturated `H_(2)S` solution to precipitate `Pb^(2+)` but not `Zn^(2+)` from `[K_(sp)H_(2)S=1.1xx10^(-22) and K_(sp)ZnS=1.0xx0^(-21)]`

Answer» `K_(sp) ` for `ZnA = 10^(-21) and [Zn^(2+)] = 0.01 M = 10^(-2) M`
But `K_(sp) [ZnS) = [ Zn^(2+) ] [S^(-2)], i.e., 10^(-21) = [10^(-2)] [S^(2-)] or [S^(2-) ] = 10^(-19) M`
Thus, to prevent precipitation of `Zn^(2+)` ions, `[S^(2-)]` must be less than `10^(19)` M. Futher,
`H_(2)S overset(2H_(2)O)rarr 2 H_(3)O^(+) + S^(2-)`
`:. K_(sp) (H_(2)S) = [ 2 H_(3)O^(+)]^(2) [S^(2-)]`
`1.1xx10^(-22)=[2H_(3)O^(+)]^(2)xx10^(-19) or [2 H_(3)O^(+)]^(2)=1.1xx10^(-3) = 11xx10^(-4)`
or `[2H_(3)O^(+)] = 3.32xx10^(-2) M or [H_(3)O^(+) ] = 1.66xx10^(-2) M`


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