The equation for the dissociation of a weak acid HA in solution is:
HA + H_{2}O ⇌ H_{3}O^{+} + A¯
and its K_{a} expression is:
[H_{3}O^{+}] [A¯] K_{a} = ––––––––– [HA]
The equation for the ionization of a weak base A¯ in solution is:
A¯ + H_{2}O ⇌ HA + OH¯
and its K_{b} expression is:
[HA] [OH¯] K_{b} = ––––––––– [A¯]
Rearrange the K_{a} expression above as follows:
[H_{3}O^{+}] [HA] –––––– = –––––– K_{a} [A¯]
and substitute the left-hand portion into the K_{b} expression from above to obtain:
[H_{3}O^{+}] [OH¯] K_{b} = ––––––––––– K_{a}
Since the equation for the ionization of water is:
K_{w} = [H_{3}O^{+}] [OH¯]
by substitution and rearrangement, we obtain:
K_{a}K_{b} = K_{w}
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 K_{a} for HA = 1.50 x 10¯^{5}, then we know that the K_{b} for A¯ (the conjugate base) must be 6.67 x 10¯^{10}. This type of calculation becomes important in doing hydrolysis and buffer calculations.