7.11.5 Relation between Ka and Kb

As seen earlier in this chapter, Ka and Kb represent the strength of an acid and a base, respectively.

In case of a conjugate acid-base pair, they are related in a simple manner so that if one is known, the other can be deduced. Considering the example of NH4+ and NH3 we see,

NH3(aq) + H2O(l) ⇔  NH3(aq) + H3O+(aq)

Ka = [H3O+][NH3]/[NH4+] = 5.6 X 10-10

NH3(aq) + H2O(l) ⇔  NH4+(aq) + OH-(aq)

Kb = [NH4+][OH-]/[NH3] = 1.8 x 10-5

Net: 2H2O(l) ⇔  H3O+(aq) + OH-(aq)

Kw = [H3O+][ OH-] = 1.0 X 10-14 M

Where, Ka represents the strength of NH4+ as an acid and Kb represents the strength of NH3 as a base.

It can be seen from the net reaction that the equilibrium constant is equal to the product of equilibrium constants Ka and Kb for the reactions added. Thus,

Ka X Kb  = {[H3O+][NH3]/[NH4+]} X {[NH4+][OH-]/[NH3]}

= [H3O+][ OH-] = Kw

=(5.6 X 10-10 ) X (1.8 x 10-5) = 1.0 X 10-14

This can be extended to make a generalisation. The equilibrium constant for a net reaction obtained after adding two (or more) reactions equals the product of the equilibrium constants for individual reactions:

KNET = K1 X K2 X…….

Similarly, in case of a conjugate acid-base pair,

Ka X Kb  = Kw

Knowing one, the other can be obtained. It should be noted that a strong acid will have a weak conjugate base and vice-versa.

Alternatively, the above expression Kw = Ka × Kb, can also be obtained by considering the base-dissociation equilibrium reaction:

B(aq) + H2O(l) ⇔  BH+(aq) + OH-(aq)

Kb= [BH+][ OH-]/ [B]

As the concentration of water remains constant it has been omitted from the denominator and incorporated within the dissociation constant. Then multiplying and dividing the above expression by [H+], we get:

Kb= [BH+][OH-][H+] /[B][H+]

= {[OH-][H+]}{[BH+]/[B][H+]}

= Kw/ Ka

Ka X Kb  = Kw

It may be noted that if we take negative logarithm of both sides of the equation, then pK values of the conjugate acid and base are related to each other by the equation:

pKa X pKb  = pKw = 14 (at 298K)

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