Calculate the solubility of A_{2}X_{3} in pur | vedclass

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(A) The dissociation of $A_{2}X_{3}$ is given by: $A_{2}X_{3} \rightarrow 2A^{3+} + 3X^{2-}$The solubility product expression is: $K_{sp} = [A^{3+}]^{

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$1.0 	imes 10^{-5} \ mol/L$

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(A) The dissociation of $A_{2}X_{3}$ is given by: $A_{2}X_{3} ightarrow 2A^{3+} + 3X^{2-}$The solubility product expression is: $K_{sp} = [A^{3+}]^{2} [X^{2-}]^{3} = 1.1 	imes 10^{-23}$Let $S$ be the solubility of $A_{2}X_{3}$. Then,$[A^{3+}] = 2S$ and $[X^{2-}] = 3S$.Substituting these into the $K_{sp}$ expression: $K_{sp} = (2S)^{2} (3S)^{3} = 4S^{2} 	imes 27S^{3} = 108S^{5}$Equating to the given value: $108S^{5} = 1.1 	imes 10^{-23}$$S^{5} = \frac{1.1 	imes 10^{-23}}{108} \approx 1.0185 	imes 10^{-25} \approx 1.0 	imes 10^{-25}$Taking the fifth root: $S = (1.0 	imes 10^{-25})^{1/5} = 1.0 	imes 10^{-5} \ mol/L$.

Calculate the solubility of $A_{2}X_{3}$ in pure water,assuming that neither kind of ion reacts with water. The solubility product of $A_{2}X_{3}$,$K_{sp} = 1.1 \times 10^{-23}$.

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