The solubility product of AB_2 is 8 times 10^ | vedclass

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(C) Let the solubility of $AB_2$ be $S \ M$.The dissociation reaction is: $AB_2 \rightleftharpoons A^{2+} + 2B^-$.Given the degree of dissociation $\a

Vedclass · Accepted Answer

$0.034$

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(C) Let the solubility of $AB_2$ be $S \ M$.The dissociation reaction is: $AB_2 ightleftharpoons A^{2+} + 2B^-$.Given the degree of dissociation $\alpha = 0.80$.Concentration of $A^{2+} = \alpha S = 0.8S$.Concentration of $B^- = 2 \alpha S = 2 	imes 0.8S = 1.6S$.$K_{sp} = [A^{2+}][B^-]^2$.$8 	imes 10^{-5} = (0.8S)(1.6S)^2$.$8 	imes 10^{-5} = 0.8S 	imes 2.56S^2$.$8 	imes 10^{-5} = 2.048S^3$.$S^3 = \frac{8 	imes 10^{-5}}{2.048} \approx 3.906 	imes 10^{-5} = 39.06 	imes 10^{-6}$.$S = \sqrt[3]{39.06 	imes 10^{-6}} \approx 3.39 	imes 10^{-2} \ M \approx 0.034 \ M$.

The solubility product of $AB_2$ is $8 \times 10^{-5} \ M^3$. If the salt is $80 \ \%$ dissociated in a saturated solution,find the solubility of the salt in $M$.

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