The solubility product of Lead sulphate,PbSO_ | vedclass

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(B) The dissociation of $PbSO_4$ is given by: $PbSO_4(s) \rightleftharpoons Pb^{2+}(aq) + SO_4^{2-}(aq)$.Let the solubility be $S \ mol \ L^{-1}$.The

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$3.45 	imes 10^{-2} \ g \ L^{-1}$

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(B) The dissociation of $PbSO_4$ is given by: $PbSO_4(s) ightleftharpoons Pb^{2+}(aq) + SO_4^{2-}(aq)$.Let the solubility be $S \ mol \ L^{-1}$.The solubility product $K_{sp} = [Pb^{2+}][SO_4^{2-}] = S 	imes S = S^2$.Given $K_{sp} = 1.3 	imes 10^{-8}$.$S = \sqrt{1.3 	imes 10^{-8}} \approx 1.14 	imes 10^{-4} \ mol \ L^{-1}$.To convert solubility from $mol \ L^{-1}$ to $g \ L^{-1}$,multiply by molar mass $(303 \ g \ mol^{-1})$:Solubility in $g \ L^{-1} = 1.14 	imes 10^{-4} \ mol \ L^{-1} 	imes 303 \ g \ mol^{-1} = 3.45 	imes 10^{-2} \ g \ L^{-1}$.

The solubility product of Lead sulphate,$PbSO_4$ is $1.3 \times 10^{-8}$. Calculate its solubility in pure water. The molecular mass of $PbSO_4 = 303 \ g \ mol^{-1}$.

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