The concentration of suphide ion in `0.1 M HCl` solution saturated with hydrogen sulphide is `1.0xx10^(-19)M`. If `10 mL` of this is added to `5 mL` of `0.04 M` solution of the following: `FeSO_(4), MnCl_(2), ZnCl_(z)` and `CdCl_(2)`. In which of these solutions precipitation will take place?

Explanation

Precipitation will occur in the solution in which ionic products is greater than solubility product , `K_(sp)`.
10 mL of solution containing `S^(2-)` ions is mixed with 5 mL of metal salt solution so that total volume becomes 15 mL.
`[S^(2-)]=1.0xx10^(-19)xx10/15=6.67xx10^(-20)`
`[Fe^(2+)]=[Mn^(2+)]=[Zn^(2+)]=[Cd^(2+)]`
`=5/15xx0.04=1.33xx10^(-2)` M
Ionic product for each of these will be
`=[M^(2+)][S^(2-)]`
`=(1.33xx10^(-2))xx(6.67xx10^(-20))`
`=8.87xx10^(-22)`
Since ionic product is greater than the `K_(sp)` of ZnS and CdS, therefore , `ZnCl_2` and `CdCl_2` solution will form precipitates.