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 solution is added to 5 mL of 0·04M solution of FeSO 4 , MnCl 2 , ZnCl 2 and CaCl 2 , in which solutions precipitation will take place? Given K sp for FeS = 6·3 × 10 –18 , MnS = 2·5 × 10 –13 , ZnS =1·6 × 10 –24 and CdS = 8·0 × 10 –27 . - Wired Faculty

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The concentration of sulphide ion in 0·1M HCl solution saturated with hydrogen sulphide is 1·0 × 10^–19M. If 10 mL of this solution is added to 5 mL of 0·04M solution of FeSO₄, MnCl₂, ZnCl₂ and CaCl₂, in which solutions precipitation will take place? Given K_sp for FeS = 6·3 × 10^–18, MnS = 2·5 × 10^–13, ZnS =1·6 × 10^–24and CdS = 8·0 × 10^–27.

Solution

Precipitation will occur in the solution for which ionic product is greater than solubility product. As 10 mL of solution containing S^2– ion is mixed with 5mL of metal salt solution, therefore,

After space mixing comma space open square brackets straight S to the power of 2 minus end exponent close square brackets space equals space left parenthesis 1.0 space cross times space 10 to the power of negative 19 end exponent straight M right parenthesis space cross times space fraction numerator left parenthesis 10 space mL right parenthesis over denominator left parenthesis 15 space mL right parenthesis end fraction space equals space 6.67 space cross times space 10 to the power of negative 20 end exponent straight M After space mixing comma space open square brackets straight M to the power of 2 plus end exponent close square brackets space equals space left parenthesis 0.04 straight M right parenthesis space cross times space fraction numerator left parenthesis 5 space mL right parenthesis over denominator left parenthesis 15 space mL right parenthesis end fraction space equals space 1.33 space cross times space 10 to the power of negative 2 end exponent straight M Ionic space product space space equals space open square brackets straight M to the power of 2 plus end exponent close square brackets space open square brackets straight S to the power of 2 minus end exponent close square brackets equals space left parenthesis 1.33 space cross times space 10 to the power of negative 2 end exponent right parenthesis space cross times space left parenthesis 6.67 space cross times space 10 to the power of negative 20 end exponent right parenthesis equals space 8.87 space cross times space 10 to the power of negative 22 end exponent space space space

Since the ionic product is greater than the solubility of ZnS and CdS, therefore both ZnCl₂and CdCl₂ solutions will be precipitated by passing H₂S gas in the presence of HCl.

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Equilibrium

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