M_{3}A_{2} is a sparingly soluble salt of mol | vedclass

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Question

(A) The molar solubility $S$ is given by $S = \frac{x}{y} 	ext{ mol L}^{-1}$.For the dissociation of the salt: $M_{3}A_{2} ightleftharpoons 3M^{2+}

Vedclass · Accepted Answer

$\frac{1}{54} \cdot \frac{y^{4}}{x^{4}}$

Vedclass · Answer

(A) The molar solubility $S$ is given by $S = \frac{x}{y} 	ext{ mol L}^{-1}$.For the dissociation of the salt: $M_{3}A_{2} ightleftharpoons 3M^{2+} + 2A^{3-}$.The solubility product $K_{sp}$ is calculated as $K_{sp} = [3S]^{3} \cdot [2S]^{2} = 27S^{3} \cdot 4S^{2} = 108S^{5}$.The molar concentration of the anion is $[A^{3-}] = 2S = 2 \left( \frac{x}{y} ight)$.The required ratio is $\frac{[A^{3-}]}{K_{sp}} = \frac{2S}{108S^{5}} = \frac{1}{54S^{4}}$.Substituting $S = \frac{x}{y}$, we get $\frac{1}{54(x/y)^{4}} = \frac{y^{4}}{54x^{4}}$.

$M_{3}A_{2}$ is a sparingly soluble salt of molar mass $y \text{ g mol}^{-1}$ and solubility $x \text{ g L}^{-1}$. The ratio of the molar concentration of the anion $(A^{3-})$ to the solubility product of the salt is

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