(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/$L$. At equilibr

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$S = (\frac{K_{sp}}{256})^{1/5}$

(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/$L$. At equilibrium,$[M^{4+}] = S$ and $[X^-] = 4S$. The solubility product $K_{sp}$ is defined as: $K_{sp} = [M^{4+}][X^-]^4$ $K_{sp} = (S)(4S)^4$ $K_{sp} = S \times 256S^4$ $K_{sp} = 256S^5$ Therefore,$S^5 = \frac{K_{sp}}{256}$ $S = (\frac{K_{sp}}{256})^{1/5}$

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

Medium

For a sparingly soluble salt $MX_4$,the molar solubility is $S$ (mol/$L$) and its solubility product is $K_{sp}$. Determine the relationship between $K_{sp}$ and $S$.

A
$S = (256 \ K_{sp})^{1/5}$
B
$S = (128 \ K_{sp})^{1/4}$
C
$S = (\frac{K_{sp}}{128})^{1/4}$
D
$S = (\frac{K_{sp}}{256})^{1/5}$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product

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For a sparingly soluble salt $MX_4$,the molar solubility is $S$ (mol/$L$) and its solubility product is $K_{sp}$. Determine the relationship between $K_{sp}$ and $S$.

Similar Questions

Evaporation of $1 \ L$ of a saturated solution of $CaCO_3$ leaves a residue of $7.0 \ g$. What is the solubility product $(K_{sp})$ for $CaCO_3$?

The $K_{sp}$ of $Mg(OH)_2$ is $1.2 \times 10^{-11}$. Calculate its solubility in pure water.

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The molar solubility of $Zn(OH)_{2}$ in $0.1 \, M \, NaOH$ solution is $x \times 10^{-18} \, M$. The value of $x$ is ...... . (Nearest integer)
(Given : The solubility product of $Zn(OH)_{2}$ is $2 \times 10^{-20}$)