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The degree of hydrolysis of 0.1 M solution of conjugate base of HA is 0.01.
Find the H⁺ concentration in 0.4 M solution of A^- :

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5 x 10^-12 M

2 x 10^-4 M

5 x 10^-11 M

2 x 10^-3 M

answer is B.

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Detailed Solution

$K_{h} = c h^{2} = 0.1 \times 10^{- 4} = 10^{- 5}$

$\begin{array}{l} 10^{- 5} = \frac{{[{OH}^{-}]}^{2}}{0.4} \\ \Rightarrow [{OH}^{-}] = 2 \times 10^{- 3} \end{array}$

or,

$[H^{+}] = 5 \times 10^{- 12}$

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