When `H_(2)S` gas is passed through an acidified solution of `Cu^(2+) and Zn^(2+)`, only CuS is precipitated-why?

Explanation

`H_(2)S` is a very weak acid. In its aqueous solution, only a small fraction of it ionises to produce `H_(3)O^(+) and S^(2-)` ions.
`[H_(2)S(aq)+2H_(2)O(l) hArr 2H_(3)O^(+)(aq)+S^(2-)(aq)]`
In an acidified solution of `H_(2)S`, due to the common ion `(H_(3)O^(+))` effect, the ionisation of `H_(2)S` is further reduced, and as a result the concentration of `S^(2-)` ions becomes extremely low.
When `H_(2)S` gas is passed through an acidified solution of `Cu^(2+)` and `Zn^(2+)` ions, the concentration of `S^(2-)` ions becomes so low that only the product of the concentrations of `Cu^(2+)` and `S^(2-)` ions exceed the solublity product of CuS. however, the product of the concentrations of `Zn^(2+) and S^(2-)` ions lies well below the value of solublity product of ZnS. this is why only CuS is precipitated in preference to ZnS.