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)
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

Category: Ionic Equilibria

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