A monoprotic acid is 0.001 % ionized in 0.1 M of its solution. The ionization constant of the acid is

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$10^{- 8} M$

$10^{- 13} M$

$10^{- 11} M$

$10^{- 5} M$

answer is C.

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Detailed Solution

$\begin{aligned} HA ⇌ H^{+} + A^{-} \\ c (1 - α) c α c α \end{aligned} K_{a} = \frac{[H^{+}] [A^{-}]}{[HA]} K_{a} = \frac{(c α)^{2}}{c (1 - α)} ≃ c α^{2}$

Hence, $K_{a} = (0.1 M) {(10^{- 5})}^{2} = 10^{- 11} M$

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