The equation for the dissociation of a weak acid HA in solution is:
HA + H2O ⇌ H3O+ + A¯
and its Ka expression is:
[H3O+] [A¯] Ka = ––––––––– [HA]
The equation for the ionization of a weak base A¯ in solution is:
A¯ + H2O ⇌ HA + OH¯
and its Kb expression is:
[HA] [OH¯] Kb = ––––––––– [A¯]
Rearrange the Ka expression above as follows:
[H3O+] [HA] –––––– = –––––– Ka [A¯]
and substitute the left-hand portion into the Kb expression from above to obtain:
[H3O+] [OH¯] Kb = ––––––––––– Ka
Since the equation for the ionization of water is:
Kw = [H3O+] [OH¯]
by substitution and rearrangement, we obtain:
KaKb = Kw
What this means is that, if we know one value for a given conjugate acid base pair, then we can calculate the value for the other member of the pair. For example, if the Ka for HA = 1.50 x 10¯5, then we know that the Kb for A¯ (the conjugate base) must be 6.67 x 10¯10. This type of calculation becomes important in doing hydrolysis and buffer calculations.