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JEE Main 20206 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2020 · 6 September · Evening Shift · Chemistry · Ionic Equilibrium

If the solubility product of AB _ 2 is 3 . 20 × 10 ^ - 11 M ^ 3 ,...

If the solubility product of AB2{AB}_{2} is 3.20×10−11M33 . 20 \times {10}^{- 11}{M}^{3}, then the solubility of AB2{AB}_{2} in pure water is _______ ×10−4molL−1\times {10}^{- 4}mol{L}^{- 1}

[Assuming that neither kind of ion reacts with water]

Correct answer

0

Step-by-step solution

{AB2⇌A2++2B−s2s\begin{cases}{AB}_{2} & ⇌ & {A}^{2 +} & + & 2{B}^{-} \\ s & 2s\end{cases}

Ksp=4s3=3.20×10−11{K}_{sp} = 4{s}^{3} = 3 .20 \times {10}^{- 11}

So solubility =2×10−4molL−1= 2 \times {10}^{- 4}mol{L}^{- 1}

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