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JEE Advanced 2018Paper 1
Exam
JEE Advanced
Year
2018
Paper
Paper 1
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Advanced 2018 · Paper 1 · Chemistry · Ionic Equilibrium

The solubility of a salt of weak acid (AB)atpH3isY × 10 ^ - 3 mol L...

The solubility of a salt of weak acid (AB)atpH3isY×10−3molL−1\left(AB\right)atpH3isY \times {10}^{- 3}mol{L}^{- 1} . The value of YY is ____. (Given that the value of solubility product of AB(Ksp)=2×10−10AB\left({K}_{sp}\right) = 2 \times {10}^{- 10}, and the value of ionization constant of HB(Ka)=1×10−8HB\left(Ka\right) = 1 \times {10}^{- 8})

Correct answer

0

Step-by-step solution

S=Ksp([H+]Ka+1)=2×10−10(10−310−8+1)≃2×10−5S = \sqrt{{K}_{sp}\left(\dfrac{\left[{H}^{+}\right]}{{K}_{a}} + 1\right)} = \sqrt{2 \times {10}^{- 10}\left(\dfrac{{10}^{- 3}}{{10}^{- 8}} + 1\right)} ≃ \sqrt{2 \times {10}^{- 5}}
=4.47×10−3M= 4 .47 \times {10}^{- 3}M.
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