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?

Answer

Precipitation will take place in cadmium chloride solution.

Explanation

For precipitation to take place, it is required that the calculated ionic product exceeds the `K_(sp)` value.
Before mixing:
`{:([S^(2-)]=1.0xx10^(-19)M,[M^(2+)]=0.04M),("volume"=10mL,"volume"=5mL):}`
After mixing:
`{:([S^(2-)]=?,[M^(2+)]=?),("volume"=(10+5)=15mL,"volume"=15mL):}`
`[S^(2-)] = (1.0 xx 10^(19) xx 10)/(15) = 6.67 xx 10^(-20) M`
`[M^(2+)] = (0.04 xx 5)/(15) = 1.33 xx 10^(-2) M`
Ionic product `= [M^(2+)][S^(2-)]`
`= (1.33 xx 10^(-2))(6.67 xx 10^(-20))`
`= 8.87 xx 10^(-22)`
This ionic product exceeds the `K_(sp)` of Zns and CdS. Therefore, precipitation will occur in `CdCl_(2)` and `ZnCl_(2)` solutions.