Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductMedium2 minPYQ_2021

ChemistryMediumstatement

$A_{3} B_{2}$ is a sparingly soluble salt of molar mass $M (g {mol}^{- 1})$ and solubility $x g L^{- 1} .$ The solubility product satisfies $K_{sp} = a {(\frac{x}{M})}^{5} .$ The value of $a$ is __________ . (Integer answer)

Answer:

108

Solution:

Solubility $= (\frac{x}{m})$ mole/lit
$A_{3} B_{2} (s) \underset{\frac{3 x}{M}}{⇌ 3 A^{2 +}} \underset{2 (\frac{x}{M})}{(aqs) + 2 B^{3 -} (aqs)}$
$K_{sp} = {(\frac{3 x}{M})}^{3} {(\frac{2 x}{M})}^{2}$
$= (3)^{3} {(2)}^{2} {(\frac{x}{m})}^{5}$
$= 27 \times 4 {(\frac{x}{m})}^{5}$
$= 108 {(\frac{x}{m})}^{5} = a {(\frac{x}{M})}^{5}$
So $a = 108$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2021

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