The solubility product of a sparingly soluble | vedclass

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Question

(A) For a sparingly soluble salt $AX_2$,the dissociation is: $AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^{-}(aq)$.Let the solubility be $s \ mol/L$.Th

Vedclass · Accepted Answer

$2 	imes 10^{-4}$

Vedclass · Answer

(A) For a sparingly soluble salt $AX_2$,the dissociation is: $AX_2(s) ightleftharpoons A^{2+}(aq) + 2X^{-}(aq)$.Let the solubility be $s \ mol/L$.Then,$[A^{2+}] = s$ and $[X^{-}] = 2s$.The solubility product expression is: $K_{sp} = [A^{2+}][X^{-}]^2 = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 3.2 	imes 10^{-11}$.$4s^3 = 3.2 	imes 10^{-11}$.$s^3 = \frac{3.2 	imes 10^{-11}}{4} = 0.8 	imes 10^{-11} = 8 	imes 10^{-12}$.$s = \sqrt[3]{8 	imes 10^{-12}} = 2 	imes 10^{-4} \ mol/L$.

The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles/litres) is

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