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

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

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(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 $\frac{1}{54(x/y)^{4}} = \frac{y^{4}}{54x^{4}}$.

$S.No.$	$Formula$	$K_{sp}$
$1$	$PQ$	$4.0 \times 10^{-20}$
$2$	$PQ_2$	$3.2 \times 10^{-14}$
$3$	$PQ_3$	$2.7 \times 10^{-35}$

$M_{3}A_{2}$ 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

Similar Questions

The solubility products of three sparingly soluble salts are given below. What is the correct decreasing order of their molar solubility?
$S.No.$ $Formula$ $K_{sp}$
$1$ $PQ$ $4.0 \times 10^{-20}$
$2$ $PQ_2$ $3.2 \times 10^{-14}$
$3$ $PQ_3$ $2.7 \times 10^{-35}$

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