The equilibrium constant of the reaction is 10^8 times the ionic product of water at 25°c Hence (ka) of weak acid will be

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Solution

Now for weak acid and strong base
Kh = Kw/Ka
Given Kh= 10^8 × Kw
So
Kw × 10^8 = Kw ÷ Ka
Ka = 10^(-8)

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The equilibrium constant of the reaction:

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Is $10^{8}$ times the ionic product of water at $25$ . Hence $K_{a}$ of weak acid will be :

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