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QJEEM03Ch5bb3: JEE Main 2003 Chemistry - Class 11 JEEM03Ch5bb3 Solution: The solubility in water of a sparingly soluble salt AB 2 is 1.0 …

The solubility in water of a sparingly soluble salt AB₂ is 1.0

\times

10^-5 mol L^-1. Its solubility product number will be :

Step-by-Step Verified Solution

A B_{2} ⇌ A^{+ 2} + 2 B^{-}

[A] = 1.0 \times 10^{- 5},

[B] = [2.0 \times 10^{- 5}],

K_{s p} = {[B]}^{2} [A]

= {[2 \times 10^{- 5}]}^{2} [1.0 \times 10^{- 5}]

= 4 \times 10^{- 15}