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 take place in the solution for which ionic product is greater than solubility product. As 10 mL of solution containing `S^(2-)` ion is mixed with 5 mL of metal salt solution , after mixing.
`[S^(2-)] =1.0 xx 10^(-19) xx (10)/(15) =6.67 xx 10^(-20)`
`[M^(2+)] i.e., [Fe^(2+)] =[Mn^(2+)] =[Zn^(2+)] =[Cd^(2+)] =(5)/(15) xx 0.04`
`=1.33 xx 10^(-2)M` lt brgt Ionic product of `[M^(2+)] " and "[S^(2-)] =(1.33 xx 10^(-2)) xx (6.67 xx 10^(-20)) =8.87 xx 10^(-22)`
As this is more than the `K_(sp)` of ZnS and CdS, both `ZnCI_(2) " and " CdCI_(2)` solution be precipitated.