If the solubility product of AB 2 is 3 . 20 × 10 - 11 M 3 , then the solubility of AB 2 in pure water is _______ × 10 - 4 mol L - 1 [Assuming that neither kind of ion reacts with water]

AB 2 ⇌ A 2 + + 2 B − s 2 s K sp = 4 s 3 = 3 .20 × 10 − 11 So solubility = 2 × 10 − 4 mol L − 1

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductEasy2 minPYQ_2020

ChemistryEasystatement

If the solubility product of ${AB}_{2}$ is $3.20 \times 10^{- 11} M^{3}$ , then the solubility of ${AB}_{2}$ in pure water is _______ $\times 10^{- 4} mol L^{- 1}$

[Assuming that neither kind of ion reacts with water]

Answer:

Solution:

$\begin{matrix} {AB}_{2} & \overset{}{⇌} & A^{2 +} & + & 2 B^{-} \\ s & 2 s \end{matrix}$

$K_{sp} = 4 s^{3} = 3 .20 \times 10^{- 11}$

So solubility $= 2 \times 10^{- 4} mol L^{- 1}$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2020

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