The solubility of a sparingly soluble salt AX | vedclass

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Question

(B) For a sparingly soluble salt of the type $AX_2$,the dissociation equilibrium is given by: $AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^-(aq)$ If $s

Vedclass · Accepted Answer

$4 	imes 10^{-12}$

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(B) For a sparingly soluble salt of the type $AX_2$,the dissociation equilibrium is given by: $AX_2(s) ightleftharpoons A^{2+}(aq) + 2X^-(aq)$ If $s$ is the solubility of the salt in $mol \ dm^{-3}$,then the concentration of $A^{2+}$ is $s$ and the concentration of $X^-$ is $2s$. The solubility product expression is: $K_{sp} = [A^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3$ Given $s = 1 	imes 10^{-4} \ mol \ dm^{-3}$. Substituting the value of $s$: $K_{sp} = 4 	imes (1 	imes 10^{-4})^3 = 4 	imes 10^{-12}$ Therefore,the correct option is $B$.

The solubility of a sparingly soluble salt $AX_2$ is $1 \times 10^{-4} \ mol \ dm^{-3}$ at $298 \ K$. Calculate its solubility product $(K_{sp})$?

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