Why is the conjugate base of a strong acid weak?

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Why is the conjugate base of a strong acid weak?

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Answer 1 · 2014-03-26T21:04:27

The product of $K_{a}$ $K_a$ and $K_{b}$ $K_b$ for any conjugate acid/base pair is always equal to $K_{w}$ $K_w$ , the self-ionization constant for water (approximately $1 x 10^{- 14}$ $1 x 10^(-14)$ ).

Therefore, larger values of $K_{a}$ $K_a$ necessarily mean that $K_{b}$ $K_b$ must be smaller (i.e., the conjugate base of a strong acid must be a weak base.

Proof:
For any acid, $H A$ $HA$ and its conjugate base, $A^{-}$ $A^-$ at equilibrium

$H A + H_{2} O \to H_{3} O^{+} + A^{-}$ $HA + H_2O → H_3O^+ + A^-$
$K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[H A]}$ $K_a = ([H_3O^+][A^-])/([HA])$

For the base reaction:
$A^{-} + H_{2} O \to O H^{-} + H A$ $A^(-) + H_2O → OH^(-) + HA$
$K_{b} = \frac{[O H^{-}] [H A]}{[A^{-}]}$ $K_b = ([OH^-][HA])/([A^-])$

$K_{a} \cdot K_{b} = [H_{3} O^{+}] [O H^{-}] = K_{w}$ $K_a · K_b = [H_3O^+][OH^-] = K_w$

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Why is the conjugate base of a strong acid weak?

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