Question

Prove that Ka x Kb = Kw for a conjugate acid/base pair.

Answer 1

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The equation of the dissociation of a weak acid HA in solution is:

HA + H₂O <===> H₃O⁺ + A¯

and its K_a 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 K_b expression is:

	[HA] [OH¯]
Kb =	----------------
	[A¯]

Rearrange the K_a expression above as follows:

[H3O+]		[HA]
----------	=	---------
Ka		[A¯]

and substitute the left-hand portion into the K_b expression from above to obtain:

	[H3O+] [OH¯]
Kb =	--------------------
	Ka

Since the equation for the ionization of water is:

K_w = [H₃O⁺] [OH¯]

by substitution and rearrangement, we obtain:

K_aK_b = K_w

_{Hope this helped.}