If the solubility product of PbS is 3.4 times | vedclass

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Question

(A) The solubility product expression for $PbS$ is given by: $K_{sp} = [Pb^{2+}][S^{2-}]$.Given that $K_{sp} = 3.4 	imes 10^{-28}$ and $[Pb^{2+}] = 0

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$3.4 	imes 10^{-25}$

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(A) The solubility product expression for $PbS$ is given by: $K_{sp} = [Pb^{2+}][S^{2-}]$.Given that $K_{sp} = 3.4 	imes 10^{-28}$ and $[Pb^{2+}] = 0.001 \ M = 10^{-3} \ M$.To initiate precipitation,the ionic product must exceed the solubility product,so we calculate the minimum concentration required: $[S^{2-}] = \frac{K_{sp}}{[Pb^{2+}]}$.Substituting the values: $[S^{2-}] = \frac{3.4 	imes 10^{-28}}{10^{-3}} = 3.4 	imes 10^{-25} \ M$.

If the solubility product of $PbS$ is $3.4 \times 10^{-28}$,what is the minimum concentration of sulfide ions required to precipitate $PbS$ from a $0.001 \ M \ Pb^{2+}$ solution?

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