Solubility of an MX_2 type electrolyte is 0.5 | vedclass

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(D) An electrolyte $MX_2$ dissociates as follows:$MX_2 \rightleftharpoons M^{2+} + 2X^{-}$If $s$ is the solubility of $MX_2$,then at equilibrium:$[M^{

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$5 	imes 10^{-13}$

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(D) An electrolyte $MX_2$ dissociates as follows:$MX_2 ightleftharpoons M^{2+} + 2X^{-}$If $s$ is the solubility of $MX_2$,then at equilibrium:$[M^{2+}] = s$$[X^{-}] = 2s$The solubility product constant $K_{sp}$ is given by:$K_{sp} = [M^{2+}][X^{-}]^2$$K_{sp} = (s)(2s)^2 = 4s^3$Given $s = 0.5 	imes 10^{-4} \ mol/L$:$K_{sp} = 4 	imes (0.5 	imes 10^{-4})^3$$K_{sp} = 4 	imes (0.125 	imes 10^{-12})$$K_{sp} = 0.5 	imes 10^{-12} = 5 	imes 10^{-13}$

Solubility of an $MX_2$ type electrolyte is $0.5 \times 10^{-4} \ mol/L$,then $K_{sp}$ of electrolyte is

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