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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 73 of 73
Marks: +1, -0
The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is 1.0 × 10−19M10^{-19} M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following : FeSO4,MnCl2,ZnCl2\mathrm{FeSO_4}, \mathrm{MnCl_2}, \mathrm{ZnCl_2} and CdCl2\mathrm{CdCl_2}, in which of these solutions precipitation will take place?
Given KspK_{sp} for
FeS = 6.3 × 10−1810^{-18}
MnS = 2.5 × 10−1310^{-13}
ZnS = 1.6 × 10−2410^{-24}
CdS = 8.0 × 10−2710^{-27}
Solution:  
Precipitation will take place in the solution for which ionic product is greater than solubility product. As 10 mL of solution containing S2−S^{2-} ion is mixed with 5 mL of metal salt solution, after mixing
[S2−][S^{2-}] = 1.0 × 10−1910^{-19} × 1015\frac{10}{15} = 6.67 × 10−2010^{-20}
[Fe2+][\mathrm{Fe}^{2+}] = [Mn2+][\mathrm{Mn}^{2+}] = [Zn2+][\mathrm{Zn}^{2+}] = [Cd+][\mathrm{Cd}^{+}] = 515\frac{5}{15} × 0.04 = 1.33 × 10−2M10^{-2} M
Hence the ionic product [M2+][S2−][M^{2+}][S^{2-}] = 1.33 × 10−210^{-2} × 6.67 × 10−2010^{-20} = 8.87 × 10−2210^{-22}
Therefore in ZnS (KspK_{sp} = 1.6 × 10−2410^{-24}) and CdS (KspK_{sp} = 8.0 × 10−2710^{-27}), ionic product exceeds solubility product. Hence, precipitation will take place.
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