InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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))` |
|
Answer» For ZnS not to be precipitated from a solution of `Zn^(2+)` and `Pb^(2+)`. `[Zn^(2+)][S^(2-)]lt K_(SP)of ZnS` `[10^(-2)][S^(2-)]lt 1.0xx10^(-21)` or themaximum `[S^(2-)]=10^(-19)` at which ZnS will begin to precipitate or upto this concentration, to precipitation will occur. `H_(2)ShArr2H^(+)+S^(2+)` `:. [H^(+)]^(2)[S^(2-)]=1.1xx10^(-22)` `:. [H^(+)]^(2) [10^(-19)]=1.1xx10^(-22)` `:. [H^(+)]^(2)=11xx10^(-4)` `:. [H^(+)]=3.3xx10^(-2)M` Thus, if `[H^(+)]=3.3xx10^(-2)` or slightly higher, the precipitation of ZnS will not take place and only PbS will precipitate. |
|