The solubility product of a salt having the g | vedclass

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Question

(A) The dissociation of the salt $MX_2$ is given by: $MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^-(aq)$.Let the solubility of the salt be $s \ mol/L$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The dissociation of the salt $MX_2$ is given by: $MX_2(s) ightleftharpoons M^{2+}(aq) + 2X^-(aq)$.Let the solubility of the salt be $s \ mol/L$. 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 $4s^3 = 4 	imes 10^{-12}$.$s^3 = 10^{-12}$,which gives $s = (10^{-12})^{1/3} = 10^{-4} \ M$.Since $[M^{2+}] = s$,the concentration of $M^{2+}$ ions is $1.0 	imes 10^{-4} \ M$.

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

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