The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitations will take place?

Given:

Concentration of sulphide ion [S2-] = 1.0 × 10–19 M


Volume of solution containing S2- ion = 10 ml


Volume of metals salt solution added = 5ml


As 10ml of solution containing S2- ion is mixed with 5ml of metals salt solution. Hence, after mixing volume will be 10ml + 5mL = 15ml


Now the concentration of sulphide ion is:


[S2-] = 1× 10-19 M × ml


⇒6.67 × 10-20


The concentration of metals solution [M2+] is


⇒[Fe2+]= [Mn2+] = [Zn2+] = [Cd2+] = × 0.04 = 1.33 × 10-2


As we know that precipitation will take place in the solution for which ionic product is greater than solubility product. Ksp < [A+][B-]


Ionic product for each of these will be [M2+][S2-]


⇒6.67 × 10-20 ×1.33 × 10-2


⇒8.87 × 10-22


As this is greater than solubility product of ZnS and CdS, therefore ZnCl2 and CdCl2 solutions will be precipitated.


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