InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A solution constains `Zn^(2+)` ions and `Cu^(2+)` ions each of `0.02M`. If the solution is made `1M` in `H^(o+)`, and `H_(2)S` is passed untill the solution is satured, should a precipitate be formed? Given: `K_(sp) ZnS = 10^(-22)`, `K_(sp) Cus = 8 xx 10^(-37)`. In satured solution, `K_(sp) (H_(2)S) = 10^(-22)` |
|
Answer» Correct Answer - A::C `H_(2)S hArr 2H^(o+) + S^(2-)` `K_(sp) = [H^(o+)]^(2) [S^(2-)]` `10^(-22) = (1M)^(2) (S^(2-):. [S^(2-)] = 10^(-22)M` i. `Q_(sp) (or IP) of ZnS = [Zn^(2+)] [S^(2-)]` `= 0.02 xx 10^(-22) = 2xx 10^(-24)M` `Q_(sp) lt K_(sp) of ZnS (2 xx 10^(-24) lt 10^(-22))`. Does not precipitate. ii. `Q_(sp) (orIP) of CuS = [Cu^(2+)] [S^(2-)]` `= 0.22 xx 10^(-22) = 2 xx 10^(-24)M` `Q_(sp) of CuS gt K_(sp) of CuS (2 xx 10^(-24) gt 8 xx 10^(-37))` So `CuS` will precipitate. |
|