Question

The concentration of sulphide ion in 0.1M $HCl$ solution saturated with hydrogen sulphide is $1.0\times {10}^{-19}M$. If 10 mL of this is added to 5 mL of 0.04 M solution of ${FeSO}_4,{MnCl}_2,{ZnCl}_2$ and ${CdCl}_2$. In which of these solutions precipitation will take place?

Answer 1

When ionic product is greater than solubility product, the precipitation will occur.
Total volume after mixing is $10+5=15$ mL.
$\left[S^{2-}\right]=\frac{1.0\times {10}^{-19}\times 10}{15}=6.67\times {10}^{-20}M$
$\left[M^{2+}\right]=\frac{5}{15}\times 0.04=0.0133M$
Ionic product will be $\left[M^{2+}\right]\left[S^{2-}\right]=0.0133\times 6.67\times {10}^{-20}=8.87\times {10}^{-22}$
In case of ZnS and CdS, the ionic product is greater than the solubility product.
Hence, $ZnC{l}_2$ and $CdC{l}_2$ solution will precipitate.