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: FeSO4, MnCl2, ZnCl2 and CdCl2 . in which of these solutions precipitation will take place?

Asked by Pragya Singh | 11 months ago |  208

##### Solution :-

If the ionic product exceeds the Ksp value, then only precipitation can take place.

Before mixing:[S2-]

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

[M2+] = 0.04M

Volume = 10mL

Volume = 5mL

After mixing:[S2-] = ?

and    [M2+] = ?

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 Ksp value. Thus, precipitation will occur in  ZnCl2 and CdCl2 solutions.

Answered by Abhisek | 11 months ago

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