What [H_(3)O^(+)] must be maintained in a satured H_(2)S solutions of precipitate Pb^(2+) but not Zn^(2+) form a solution in which each ion is present at a concentration of 0.01M? (K_(sp) of H_(2)S=1.1xx10^(-22),K_(sp) of ZnS=1.0xx10^(-21),sqrt(11)=3.3)

What [H_(3)O^(+)] must be maintained in a satured H_(2)S solutions of

1.	What [H_(3)O^(+)] must be maintained in a satured H_(2)S solutions of precipitate Pb^(2+) but not Zn^(2+) form a solution in which each ion is present at a concentration of 0.01M? (K_(sp) of H_(2)S=1.1xx10^(-22),K_(sp) of ZnS=1.0xx10^(-21),sqrt(11)=3.3)
Answer» Solution :For `ZnS` not to be precipitate from a solution of `Zn^(2+)` and `Pb^(2+),IP_(ZnS)ltK_(sp(ZnS))` `[Zn^(2+)][S^(2-)] ltK_(sp) of ZnS` `[10^(-2)][S^(2-)] lt1.0xx10^(-21):.[S^(2-)lt10^(-19)M` so at `[S^(2-)]=10^(-19)M` or less no PRECIPITAION of `ZnS` will occur. `H_(2)S HARR 2H^(+)+S^(2-)` `:.[H^(+)]^(2)[S^(2-)]=K_(sp(H_(2)S))=1.1xx10^(-22):.[H^(+)]_(min)^(2)=1.1xx10^(-22)` `[H^(+)]_(min)^(2)=11xx10^(-4):.[H^(+)]_(min)=3.3xx10^(-22)` Thus if `[H^(+)=3.3xx10^(-2)M` or more the precipitaion of `ZnS` will not take PLACE and only `PbS` will precipitate .