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The $pH$ of a 0.1 molar solution of the acid $HQ$ is 3. The value of the ionization constant $K_{a}$ of this acid is

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$3 \times 10^{- 1} M$

$1 \times 10^{- 3} M$

$1 \times 10^{- 5} M$

$1 \times 10^{- 7} M$

answer is C.

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Detailed Solution

For a weak acid, we have $HQ ⇌ H^{+} + Q^{-}$ ; $K_{a} = \frac{[H^{+}] [Q^{-}]}{[HQ]}$

Also $[H^{+}] ≃ [Q^{-}]$ and $[HQ] = [{HQ}_{0}$

Thus, $K_{a} = \frac{{[H^{+}]}^{2}}{[HQ]_{0}} = \frac{{(1 \times 10^{- 3} M)}^{2}}{0.1 M} = 1 \times 10^{- 5} M$

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