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(B) For a salt of type $MX_2$,the dissociation is given by: $MX_2(s) ⇌ M^{2+}(aq) + 2X^-(aq)$.Let the solubility be $S \ M$. Then,$[M^{2+}] = S$ and $

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$1.0 	imes 10^{-4} \ M$

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(B) For a salt of type $MX_2$,the dissociation is given by: $MX_2(s) ⇌ M^{2+}(aq) + 2X^-(aq)$.Let the solubility be $S \ M$. Then,$[M^{2+}] = S$ and $[X^-] = 2S$.The solubility product expression is $K_{sp} = [M^{2+}][X^-]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 4 	imes 10^{-12}$,we have $4S^3 = 4 	imes 10^{-12}$.$S^3 = 10^{-12}$,so $S = \sqrt[3]{10^{-12}} = 10^{-4} \ M$.Since $[M^{2+}] = S$,the concentration of $M^{2+}$ ions is $1.0 	imes 10^{-4} \ M$.

The solubility product of a salt having general formula $MX_2$ in water is $4 \times 10^{-12}$. The concentration of $M^{2+}$ ions in the aqueous solution of the salt is

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$MX$ is a salt formed by neutralization of a strong base,$MOH$,and a weak acid,$HX$. If the dissociation constant of $HX$ is $K_a$ and the solubility product of $MX$ is $K_{sp}$,then the solubility of $MX$ in an aqueous acidic solution is given by:

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