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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For a salt $MX_2$,the dissociation equilibrium is: $MX_2(s) ightleftharpoons M^{2+}(aq) + 2X^-(aq)$.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 $K_{sp} = 4.0 	imes 10^{-12}$,we have $4s^3 = 4.0 	imes 10^{-12}$.$s^3 = 1.0 	imes 10^{-12}$.$s = (1.0 	imes 10^{-12})^{1/3} = 1.0 	imes 10^{-4} \ M$.Since $[M^{2+}] = s$,the concentration of $M^{2+}$ is $1.0 	imes 10^{-4} \ M$.

$S.No.$	$Formula$	$K_{sp}$
$1$	$PQ$	$4.0 \times 10^{-20}$
$2$	$PQ_2$	$3.2 \times 10^{-14}$
$3$	$PQ_3$	$2.7 \times 10^{-35}$

The solubility product of a salt with the general formula $MX_2$ is $4.0 \times 10^{-12}$. What is the concentration of the $M^{2+}$ ion in a saturated solution of this salt?

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