Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductMedium2 minQB

ChemistryMediumsingle choice

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

Options:

Answer:

Solution:

Solubility $S = \frac{x}{M}$

$A_{3} B_{2} (s) ⇌= 3 A^{2 -}$ (aq.) $+ 2 s 2 B^{3 -}$ (aq.)

$K_{sp} = (3 S)^{3} (2 S)^{2} = 108 S^{5}$

$\frac{[ B ^{3 -} ]}{K _{sp}} = \frac{2 S}{108 S ^{5}} = \frac{1}{54} \frac{M ^{4}}{x ^{4}}$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: QB

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