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0.2 millimoles of ${Zn}^{2 +}$ ion is mixed with ${({NH}_{4})}_{2}$ S of molarity 0.02 M. The amount of ${Zn}^{2 +}$ that remains unprecipitated in 20 mL of this solution would be : (Given: $K_{sp}$ (ZnS) = 4 x 10- 24 )

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$2 \times 10^{- 23} g$

$5.2 \times 10^{- 22} g$

none of these

$2.6 \times 10^{- 22} g$

answer is A.

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Detailed Solution

$\underset{\underset{0}{0.2}}{{Zn}^{2 +}} + \underset{0.002 x \underset{0.2}{20} = 0.04}{S^{2 -}} ⟶ ZnS$

$\begin{array}{l} [{Zn}^{2 +}] \times (\frac{0.2}{20}) = 4 \times 10^{- 24} \\ [{Zn}^{2 +}] = 4 \times 10^{- 22} \end{array}$

$\begin{array}{l} Mass = 4 \times 10^{- 22} \times \frac{20}{1000} \times 65 \\ = 5.2 \times 10^{- 22} g \end{array}$

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