The solubility product of lead bromide is 8 t | vedclass

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(D) Let the solubility of $PbBr_2$ be $S \ M$.The dissociation reaction is: $PbBr_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Br^-_{(aq)}$.Given the d

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$3.4 	imes 10^{-2} \ M$

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(D) Let the solubility of $PbBr_2$ be $S \ M$.The dissociation reaction is: $PbBr_{2(s)} ightleftharpoons Pb^{2+}_{(aq)} + 2Br^-_{(aq)}$.Given the degree of dissociation $\alpha = 0.8$,the concentrations of ions in the saturated solution are:$[Pb^{2+}] = \alpha S = 0.8S$$[Br^-] = 2 \alpha S = 2 	imes 0.8S = 1.6S$The solubility product expression is $K_{sp} = [Pb^{2+}][Br^-]^2$.Substituting the values: $8 	imes 10^{-5} = (0.8S) 	imes (1.6S)^2$.$8 	imes 10^{-5} = 0.8S 	imes 2.56S^2 = 2.048S^3$.$S^3 = \frac{8 	imes 10^{-5}}{2.048} \approx 3.906 	imes 10^{-5} = 39.06 	imes 10^{-6}$.$S = \sqrt[3]{39.06} 	imes 10^{-2} \approx 3.39 	imes 10^{-2} \ M \approx 3.4 	imes 10^{-2} \ M$.

The solubility product of lead bromide is $8 \times 10^{-5}$. If the salt is $80\%$ dissociated in a saturated solution,find the solubility of the salt.

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