The equilibrium constant of the reaction:
$A^{-} + H_{2} O ⇌ H A + {O H}^{-}$
Is $10^{8}$ times the ionic product of water at $25$ . Hence $K_{a}$ of weak acid will be :

10^{- 8}

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10^{- 6}

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10^{- 14}

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None of these

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Solution

The correct option is D $10^{- 8}$
$\begin{matrix} Given, k_{h} = 10^{8} k_{ω} The relation between k_{h} and k_{a} is given as : - k_{h} = \frac{k_{ω}}{k_{a}} \Rightarrow 10^{8} k_{ω} = \frac{k_{ω}}{k_{a}} \Rightarrow k_{a} = 10^{- 8} \end{matrix}$

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