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

Question

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

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)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

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 .