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 0.04M solution of the following: FeSO_{4}, MnCl_{2}, ZnCl_{2} and CdCl_{2} . in which of these solutions precipitation will take place?

Asked by Pragya Singh | 1 year ago | 227

If the ionic product exceeds the K_{sp} value, then only precipitation can take place.

Before mixing:[S^{2-}]

=\( K_{sp} = 1.0\times 10^{-19}\)

[M^{2+}] = 0.04M

Volume = 10mL

Volume = 5mL

After mixing:[S^{2-}] = ?

and [M^{2+}] = ?

Total volume = (10 + 5) = 15mL

Volume = 15mL

\( [S^{2-}] = \dfrac{1.0\times 10^{-19}\times 10}{15} \)

\( = 6.67\times 10^{-20}M[M^{2+}] \)

\( = \dfrac{0.04\times 5}{15} = 1.33\times 10^{-2}\)

Now, the ionic product

=\( [M^{2+}][S^{2-}]\)

\( = (1.33\times 10^{-2})(6.67\times 10^{-20})\)

\( = 8.87\times 10^{-22}\)

Here, the ionic product of CdS and ZnS exceeds its corresponding K_{sp} value. Thus, precipitation will occur in ZnCl_{2} and CdCl_{2} solutions.

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