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Solubility Product

A 3 B 2 is a ...

Question

$A_{3} B_{2}$ is a sparingly soluble salt of molar mass M ( ${g m o l}^{- 1})$ and solubility x g litre $^{- 1}$ . The ratio of the molar concentration of $B^{3 -}$ to the solubility product of the salt is :

108 \frac{X^{5}}{M^{5}}

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\frac{1}{108} \frac{M^{4}}{X^{4}}

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\frac{1}{54} \frac{M^{4}}{X^{4}}

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Solution

The correct option is D $\frac{1}{54} \frac{M^{4}}{X^{4}}$
Solubility $S = \frac{x}{M}$
$A_{3} B_{2} (s) ⇌ 3 A^{2 -} (a q .) 3 S + 2 B^{3 -} (a q .) 2 S$
$K_{s p} = (3 S)^{2} (2 S)^{2} = 108 S^{5}$
$\frac{[B^{3 -}]}{K_{s p}} = \frac{2 S}{108 S^{5}} = \frac{1}{54} \frac{M^{4}}{x^{4}}$

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A3B2 is a sparingly soluble salt of molar mass M (gmol−1) and solubility x g litre−1 . The ratio of the molar concentration of B3− to the solubility product of the salt is :

The correct option is D 154M4X4Solubility S=xM A3B2(s)⇌3A2−(aq.)3S+2B3−(aq.)2SKsp=(3S)2(2S)2=108S5[B3−]Ksp=2S108S5=154M4x4

$A_{3} B_{2}$ is a sparingly soluble salt of molar mass M ( ${g m o l}^{- 1})$ and solubility x g litre $^{- 1}$ . The ratio of the molar concentration of $B^{3 -}$ to the solubility product of the salt is :

The correct option is D $\frac{1}{54} \frac{M^{4}}{X^{4}}$
Solubility $S = \frac{x}{M}$
$A_{3} B_{2} (s) ⇌ 3 A^{2 -} (a q .) 3 S + 2 B^{3 -} (a q .) 2 S$
$K_{s p} = (3 S)^{2} (2 S)^{2} = 108 S^{5}$
$\frac{[B^{3 -}]}{K_{s p}} = \frac{2 S}{108 S^{5}} = \frac{1}{54} \frac{M^{4}}{x^{4}}$