InterviewSolution
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InterviewSolution
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The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is `1.0xx10^(-9)M`. If 10 mL of this solution is added to 5 mL of 0.04 M solution of `FeSO_(4), MnCl_(2), ZnCl_(2) and CdCl_(2)`, in which solutions precipitation will take place ? Given `K_(sp)` for `Fes=6.3xx10^(-18), MnS = 2.5xx10^(-13), ZnS=1.6xx10^(-24) and CdS=8.0xx10^(-27)`. |
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Answer» 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. |
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