The solubility of PbF_2 in water at 25,^{circ | vedclass

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Question

(A) The solubility of $PbF_2$ in water is $S = 10^{-3}\, M$. The solubility product constant $K_{sp}$ is given by $K_{sp} = [Pb^{2+}][F^-]^2 = (S)(2S)

Vedclass · Accepted Answer

$1.6 	imes 10^{-6}\, M$

Vedclass · Answer

(A) The solubility of $PbF_2$ in water is $S = 10^{-3}\, M$. The solubility product constant $K_{sp}$ is given by $K_{sp} = [Pb^{2+}][F^-]^2 = (S)(2S)^2 = 4S^3$. $K_{sp} = 4 	imes (10^{-3})^3 = 4 	imes 10^{-9}$. In $0.05\, M\, NaF$ solution,$NaF$ dissociates completely to provide $[F^-] = 0.05\, M$. Let the solubility of $PbF_2$ in this solution be $S'\, M$. Then $[Pb^{2+}] = S'$ and $[F^-] = (2S' + 0.05)\, M$. $K_{sp} = S'(2S' + 0.05)^2 = 4 	imes 10^{-9}$. Since $S'$ is very small,we assume $2S' \ll 0.05$,so $(2S' + 0.05) \approx 0.05$. $S' 	imes (0.05)^2 = 4 	imes 10^{-9}$. $S' 	imes 2.5 	imes 10^{-3} = 4 	imes 10^{-9}$. $S' = \frac{4 	imes 10^{-9}}{2.5 	imes 10^{-3}} = 1.6 	imes 10^{-6}\, M$.

The solubility of $PbF_2$ in water at $25\,^{\circ}C$ is $\sim 10^{-3}\, M$. What is its solubility in $0.05\, M\, NaF$ solution? Assume the latter to be fully ionised.

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