The solubility product of PbS is 3.4 times 10 | vedclass

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(B) The solubility product expression for $PbS$ is given by: $K_{sp} = [Pb^{2+}][S^{2-}]$.Given $K_{sp} = 3.4 	imes 10^{-28}$ and $[Pb^{2+}] = 1 	im

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$3.4 	imes 10^{-26}$

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(B) The solubility product expression for $PbS$ is given by: $K_{sp} = [Pb^{2+}][S^{2-}]$.Given $K_{sp} = 3.4 	imes 10^{-28}$ and $[Pb^{2+}] = 1 	imes 10^{-2} \ mol/L$.To precipitate $PbS$,the ionic product must exceed the solubility product $(Q_{sp} > K_{sp})$.Therefore,$[S^{2-}] > \frac{K_{sp}}{[Pb^{2+}]}$.$[S^{2-}] > \frac{3.4 	imes 10^{-28}}{1 	imes 10^{-2}}$.$[S^{2-}] > 3.4 	imes 10^{-26} \ mol/L$.Thus,the minimum concentration required is $3.4 	imes 10^{-26} \ mol/L$.

The solubility product of $PbS$ is $3.4 \times 10^{-28}$. If $[Pb^{2+}] = 1 \times 10^{-2} \ mol/L$,what concentration of $[S^{2-}]$ is required to precipitate $PbS$?

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