What is the solubility of PbSO_4 in centinorm | vedclass

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(A) The dissociation of $PbSO_4$ is given by: $PbSO_4(s) ⇌ Pb^{2+}(aq) + SO_4^{2-}(aq)$.Given $K_{sp} = [Pb^{2+}][SO_4^{2-}] = 1 	imes 10^{-8}$.In $0

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$2 	imes 10^{-6} \ mol/L$

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(A) The dissociation of $PbSO_4$ is given by: $PbSO_4(s) ⇌ Pb^{2+}(aq) + SO_4^{2-}(aq)$.Given $K_{sp} = [Pb^{2+}][SO_4^{2-}] = 1 	imes 10^{-8}$.In $0.01 \ N \ H_2SO_4$,the concentration of $SO_4^{2-}$ ions is determined by the normality of the acid.Since $H_2SO_4$ is a diprotic acid,$0.01 \ N \ H_2SO_4 = 0.005 \ M \ H_2SO_4$.Thus,$[SO_4^{2-}] = 0.005 \ M = 5 	imes 10^{-3} \ M$.Let the solubility of $PbSO_4$ in this solution be $s'$.Then $[Pb^{2+}] = s'$ and $[SO_4^{2-}] = (s' + 5 	imes 10^{-3}) \approx 5 	imes 10^{-3} \ M$ (since $s'$ is very small).$K_{sp} = s' 	imes (5 	imes 10^{-3}) = 1 	imes 10^{-8}$.$s' = \frac{1 	imes 10^{-8}}{5 	imes 10^{-3}} = 0.2 	imes 10^{-5} = 2 	imes 10^{-6} \ mol/L$.

What is the solubility of $PbSO_4$ in centinormal $(0.01 \ N)$ $H_2SO_4$ solution? The solubility product $(K_{sp})$ of $PbSO_4$ at a given temperature is $1 \times 10^{-8} \ (mol/L)^2$.

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