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)
Ka (HA) = [H3O+] [A–] / [HA]…………………. (1)
The conjugate base, A– behaves as a weak base in water,
A– + H2O < ------------> HA + OH–
Kb (A–) = [HA] [OH–] / [A–]…………………. (2)
Multiplying equation (1) and (2), we get
Ka (HA) x Kb (A–)
= [[H3O+] [A–] x [HA] ]/[HA] [OH–] / [A–]
= [H3O+] [OH–] = Kw
Ka (HA) x Kb (A–) = Kw
In general, for any acid
Ka (acid) x Kb (conjugate base) = Kw
= 10-14 mol2 L-2 at 298 K
Taking negative logarithm of both sides we will get:
-Log Ka (acid) x – log Kb (conjugate base) = – log Kw
pKa x pKb = pKw
At 298 K, pKw = – log 10-14 = 14
Therefore, pKa (acid) x pKb (conjugate base) = 14 at 298 K