Derive the relationship between Ka and Kb for an acid and its conjugate base.

Conjugate acid and base are related as:
Base + H⁺ <—————> acid
Acid <—————–> H⁺ + base
Let us consider a weak acid, HA
HA (aq) + H₂O < —————> H₃O⁺ (aq) + A^– (aq)
K_a (HA) = [H₃O⁺] [A^–] / [HA]…………………. (1)
The conjugate base, A^– behaves as a weak base in water,
A^– + H₂O < ————> HA + OH^–
K_b (A^–) = [HA] [OH^–] / [A^–]…………………. (2)
Multiplying equation (1) and (2), we get
K_a (HA) x K_b (A^–)
= [[H₃O⁺] [A^–] x [HA] ]/[HA] [OH^–] / [A^–]

= [H₃O⁺] [OH^–] = K_w
K_a (HA) x K_b (A^–) = K_w
In general, for any acid
K_a (acid) x K_b (conjugate base) = K_w
= 10^-14 mol² L^-2 at 298 K
Taking negative logarithm of both sides we will get:
-Log K_a (acid) x – log K_b (conjugate base) = – log K_w
pK_a x pK_b = pK_w
At 298 K, pK_w = – log 10^-14 = 14
Therefore, pK_a (acid) x pK_b (conjugate base) = 14 at 298 K