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

A, C

Explanation

`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.