InterviewSolution
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InterviewSolution
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Determine whether `Cd^(2+)` can be separed form `Zn^(2+)` by bubbling `H_(2)S` through a `0.3M` HCI solution that contgains `0.005M Cd^(2+)` and `0.005M Zn^(2+)` . `(K_(SP)` for CdS and ZnS are `8xx10^(-7)` and `3xx10^(-2))` respectively) |
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Answer» On bubbling `H_(2)S`, the solution becomes saturated with `H_(2)S` to give MS and thus `[H_(2)S]=0.005=[M^(2+)]` `M^(2+)+H_(2)S+2H_(2)OhArr MS_((s))+2H_(3)O_((aq.))^(+)` `Q_(f)=([H_(3)O^(+)]^(2))/([M^(2+)][H_(2)S])= ((0.3)^(2))/((0.005)xx(0.005))=3600` or for reverse reaction `Q_(r)= 2.77xx10^(-4)`. Since `Q_(r)gt K_(SP)` of CdS, thus CdS will be precipitated. Also `Q_(r)lt K_(SP)` of ZnS, thus, ZnS will remain in solution. |
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