### Derive this Equation: KaKb = Kw

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.