An acidified solution of 0.05 M Zn^{2+} is | vedclass

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Question

(C) To prevent the precipitation of $ZnS$,the ionic product $Q_{sp}$ must be less than or equal to the solubility product $K_{sp}$.$Q_{sp} = [Zn^{2+}]

Vedclass · Accepted Answer

$0.20$

Vedclass · Answer

(C) To prevent the precipitation of $ZnS$,the ionic product $Q_{sp}$ must be less than or equal to the solubility product $K_{sp}$.$Q_{sp} = [Zn^{2+}][S^{2-}] \leq K_{sp}(ZnS)$Given: $[Zn^{2+}] = 0.05 \ M$ and $K_{sp}(ZnS) = 1.25 	imes 10^{-22}$$[S^{2-}] \leq \frac{K_{sp}}{[Zn^{2+}]} = \frac{1.25 	imes 10^{-22}}{0.05} = 2.5 	imes 10^{-21} \ M$For the dissociation of $H_2S$: $H_2S ightleftharpoons 2H^{+} + S^{2-}$The overall dissociation constant is given by:$K_{NET} = \frac{[H^{+}]^2 [S^{2-}]}{[H_2S]}$$1 	imes 10^{-21} = \frac{[H^{+}]^2 	imes (2.5 	imes 10^{-21})}{0.1}$$[H^{+}]^2 = \frac{1 	imes 10^{-21} 	imes 0.1}{2.5 	imes 10^{-21}} = \frac{0.1}{2.5} = 0.04$$[H^{+}] = \sqrt{0.04} = 0.2 \ M$Thus,the minimum concentration of $H^{+}$ required is $0.20 \ M$.

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