`H_(2)S` is passed into one `dm^(3)` of a solution containing 0.1 mole of `Zn^(2+)` and 0.01 mole of `Cu^(2+)` till the sulphide ion concentration reaches `8.1xx10^(-19)` moles .Which one of the following statements is true?
[`K_(sp)` of ZnS and CuS are `3xx10^(-22)` and `8xx10^(-36)` respectively]

Options

A. Only ZnS precipitates
B. Both CuS and ZnS precipitate
C. Only CuS precipitates
D. No precipitation occurs

Answer

B

Explanation

Precipitation occures only when ionic product exceeds the value of solubility product.
1`dm^(3)` of a solution containing 0.1 mole of `Zn^(2+)`
0.01 mole of `Cu^(2+)` and `8.1xx10^(-19)` mole of `S^(2+)` .Let Ionic product of ZnS=`[Zn^(2+)][S^(2-)]`
`=0.1xx8.1xx10^(-19)=8.1xx10^(-20)`
`K_(sp)` of ZnS`=3xx10^(-22)`
`therefore` Ionic product`gtK_(sp)`
Both ZnS and CuS have less `K_(sp)` values than their respective ionic products so ZnS and CuS both precipitates.