A salt MX has K_{sp} = 4 times 10^{-10}. What | vedclass

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(D) For salt $MX$,the solubility product expression is $K_{sp} = s^2$,where $s$ is the molar solubility.Given $K_{sp} = 4 	imes 10^{-10}$,so $s = \sq

Vedclass · Accepted Answer

$4.32 	imes 10^{-18}$

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(D) For salt $MX$,the solubility product expression is $K_{sp} = s^2$,where $s$ is the molar solubility.Given $K_{sp} = 4 	imes 10^{-10}$,so $s = \sqrt{4 	imes 10^{-10}} = 2 	imes 10^{-5} \ mol \ L^{-1}$.For salt $MX_3$,the dissociation is $MX_3 ightleftharpoons M^{3+} + 3X^-$. The solubility product expression is $K_{sp} = (s)(3s)^3 = 27s^4$.Substituting the value of $s = 2 	imes 10^{-5} \ mol \ L^{-1}$ into the expression for $MX_3$:$K_{sp} = 27 	imes (2 	imes 10^{-5})^4 = 27 	imes 16 	imes 10^{-20} = 432 	imes 10^{-20} = 4.32 	imes 10^{-18}$.

$A$ salt $MX$ has $K_{sp} = 4 \times 10^{-10}$. What value of $K_{sp}$ must another salt $MX_3$ have if the molar solubility of the two salts is to be identical?

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