The molar solubility of a sparingly soluble s | vedclass

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(D) The dissociation of the salt $MX_4$ is given by: $MX_4(s) \rightleftharpoons M^{4+}(aq) + 4X^{-}(aq)$ Let the solubility be $s \ mol \ L^{-1}$. Th

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$s = (K_{sp}/256)^{1/5}$

Vedclass · Answer

(D) The dissociation of the salt $MX_4$ is given by: $MX_4(s) ightleftharpoons M^{4+}(aq) + 4X^{-}(aq)$ Let the solubility be $s \ mol \ L^{-1}$. Then,$[M^{4+}] = s$ and $[X^{-}] = 4s$. The solubility product expression is: $K_{sp} = [M^{4+}][X^{-}]^4$ $K_{sp} = (s)(4s)^4$ $K_{sp} = s 	imes 256s^4$ $K_{sp} = 256s^5$ Rearranging for $s$: $s^5 = \frac{K_{sp}}{256}$ $s = (\frac{K_{sp}}{256})^{1/5}$

The molar solubility of a sparingly soluble salt $MX_4$ is $s \ mol \ L^{-1}$. If the solubility product is $K_{sp}$,then $s$ in terms of $K_{sp}$ is:

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