The equilibrium constant for this net reaction is Kw = [H+][OH−], which is the product of Ka and Kb.
Thus, if the dissociation constant either for an acid or for its conjugate base is known, then the
dissociation constant for the other can be determined, using the equation:
Ka × Kb = Kw = 1 × 10−14Thus Ka and Kb are inversely related. In other words, if Ka is large (the acid is strong), then Kb will be
small (the conjugate base will be weak), and vice versa. This is a more mathematical way of
describing a general rule: The stronger an acid is, the weaker its conjugate base is (as a base); the
weaker the acid, the stronger its conjugate base. For example, HCl (Ka ~ 107 ) is a much stronger acid
than acetic acid (Ka = 1.8 × 10−5); CH 3 COO− is therefore expected to be a much stronger base than Cl−.
The mathematics should not be allowed to obscure how much sense this makes: after all, a weak
acid means that it undergoes dissociation reluctantly; the equilibrium lies to the left, favoring
undissociated HA. A− in solution is therefore likely to grab a proton to reconstitute HA; that is, A− is
reactive as a base.
APPLICATIONS OF Ka AND Kb
To calculate [H+] in a 2.0 M aqueous solution of acetic acid, first write the equilibrium reaction:
CH 3 COOH (aq) H+ (aq) + CH 3 COO− (aq)Next, write the expression for the acid dissociation constant:
The concentration of CH 3 COOH at equilibrium is equal to its initial concentration, 2.0 M, less the
amount dissociated, x. Likewise, [H+] = [CH 3 COO−] = x, since each molecule of CH 3 COOH dissociates
into one H+ ion and one CH 3 COO− ion. Thus, the equation can be rewritten as follows: