The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is `1.0 xx 10^(-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_(2)` and `CdCl_(2)`. in which of these solutions precipitation will take place?

Explanation

Precipitations takes place only when the ionic product exceeds the solubility product.
Since 10 mL of the solutions containing `S^(2-)` ions is mixed with 5mL of the different metal salt solutions ,
Volume of the mixture after mixing in each case = 10+ 5 = 15mL
` therefore" " [S^(2-) ]= 1.0 xx 10 ^(-19) xx(10)/(15) = 6.67 xx 10 ^(-20 ) M`
and ` " "[Fe^(2+) ] [ Mn^(2+)] = [Zn^(2+)] = [Cd^(2+)]`
` " " = 0.04 xx (5)/(15) = 1.33 xx10 ^(-12) M`
Therefore , in each case
Ionic product = ` [M^(2+)] [S^(2-)]`
` " " = (1.33 xx 10 ^(-2)) xx (6.67 xx 10 ^(-20))`
` " " = 8.87 xx 10 ^(-22)`
This ionic product is greater than the solubility product of Zns and CdS. Hence precipitations will take palce only in `ZnCl_2 and CdCl_2` solutions.