At 25,^oC,K_{sp} for PbBr_2 is equal to 8 tim | vedclass

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(A) Let the solubility of $PbBr_2$ be $S \, mol/L$.Given that the salt is $80\%$ dissociated,the concentration of ions in the solution will be:$[Pb^{2

Vedclass · Accepted Answer

$[\frac{10^{-4}}{1.6 	imes 1.6}]^{1/3}$

Vedclass · Answer

(A) Let the solubility of $PbBr_2$ be $S \, mol/L$.Given that the salt is $80\%$ dissociated,the concentration of ions in the solution will be:$[Pb^{2+}] = 0.8 	imes S$$[Br^-] = 2 	imes 0.8 	imes S = 1.6 	imes S$The solubility product expression is $K_{sp} = [Pb^{2+}][Br^-]^2$.Substituting the values: $8 	imes 10^{-5} = (0.8 	imes S)(1.6 	imes S)^2$.$8 	imes 10^{-5} = 0.8 	imes S 	imes 2.56 	imes S^2$.$8 	imes 10^{-5} = 2.048 	imes S^3$.$S^3 = \frac{8 	imes 10^{-5}}{2.048} = \frac{8 	imes 10^{-5}}{0.8 	imes 1.6 	imes 1.6} = \frac{10^{-4}}{1.6 	imes 1.6}$.Therefore,$S = [\frac{10^{-4}}{1.6 	imes 1.6}]^{1/3}$.

At $25\,^oC$,$K_{sp}$ for $PbBr_2$ is equal to $8 \times 10^{-5}$. If the salt is $80\%$ dissociated,what is the solubility of $PbBr_2$ in $mol/L$?

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