# Why is the conjugate base of a strong acid weak?

Mar 26, 2014

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

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

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

HA + H_2O → H_3O^+ + A^-
${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$

For the base reaction:
A^(-) + H_2O → OH^(-) + HA
${K}_{b} = \frac{\left[O {H}^{-}\right] \left[H A\right]}{\left[{A}^{-}\right]}$

K_a · K_b = [H_3O^+][OH^-] = K_w