At 25^{circ} C, the solubility product of a s | vedclass

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(B) For a salt of $MX_{2}$ type, the dissociation is given by: $MX_{2} \rightleftharpoons M^{2+} + 2X^-$. If $s$ is the solubility in $mol / L$, then

Vedclass · Accepted Answer

$2 	imes 10^{-3}$

Vedclass · Answer

(B) For a salt of $MX_{2}$ type, the dissociation is given by: $MX_{2} ightleftharpoons M^{2+} + 2X^-$. If $s$ is the solubility in $mol / L$, then $[M^{2+}] = s$ and $[X^-] = 2s$. The solubility product $K_{sp}$ is given by: $K_{sp} = [M^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3$. Given $K_{sp} = 3.2 	imes 10^{-8}$. So, $4s^3 = 3.2 	imes 10^{-8}$. $s^3 = \frac{3.2 	imes 10^{-8}}{4} = 0.8 	imes 10^{-8} = 8 	imes 10^{-9}$. Taking the cube root, $s = \sqrt[3]{8 	imes 10^{-9}} = 2 	imes 10^{-3} \ mol / L$.

At $25^{\circ} C$, the solubility product of a salt of $MX_{2}$ type is $3.2 \times 10^{-8}$ in water. The solubility (in $mol / L$) of $MX_{2}$ in water at the same temperature will be:

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