The concentration of sulphide ion in 0.01 M HCl solution saturated with hydrogen sulphide is `1.0xx10^(-19)`M. If 10 mL of this solution is added to 5 mL of 0.04 M solution of the following : `FeSO_4 , MnCl_2, ZnCl_2` and `CdCl_2`, in which of these solutions precipitation will take place ?
Given `K_(sp)`, values : `FeS=6.3xx10^(-18) , MnS=2.5xx10^(-13),ZnS=1.6xx10^(-24), CdS=8.0xx10^(-27)`

Explanation

Precipitation will take place in the solution for which ionic product is greater than solubility product. As 1o mL of solution containing `S^(2-)` ion is mixed with 5 mL of metal salt solution, after mixing
`[S^(2-)]=1.0xx10^(-19)xx(10)/(15)=6.67 xx 10^(-20)`.
`[Fe^(2+)]=[Mn^(2+)]=[Zn^(2+)]=[Cd^(2+)]=(5)/(15)xx0.04 = 1.33xx10^(-2)M`
`:.` Ionic product for each of these will be `=[M^(2+)][S^(2-)]=(1.33xx10^(-2))(6.67xx10^(-20))=8.87xx10^(-22)`
As this is greater than the solubility product of ZnS and CdS, therefore , `ZnCl_(2) and CdCl_(2)` solutions will be precipitated.