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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?

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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.

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