5.4 : 

Consider a reaction between ammonia and an ethoxide ion. Ammonia loses its proton to the ethoxide ion, forming an amide ion and ethanol.

The reaction, at equilibrium, has two acids and two bases. The stronger acid and the stronger base, along with the position of equilibrium, are identified by comparing the pK_a values of each acid.

Here, ethanol with a pK_a of 15.9 is a stronger acid than ammonia, whose pK_a is 38. Because a strong acid results in a weak conjugate base, and conversely, a weak acid provides a strong conjugate base, the ethoxide ion is a weaker base than the amide ion. 

Fundamentally, any equilibrium controlled acid–base reaction favors the formation of more stable species — the weaker acid and the weaker base — owing to their lower potential energies.

This explains why, in this reaction, the position of equilibrium lies to the side of the ethoxide ion and ammonia.

Furthermore, in any reaction, if the difference in the pKa values of the acids is large, the reversible reaction is neglected, and the equilibrium arrows are replaced by an irreversible arrow.

The position of equilibrium in an acid—base reaction can also be estimated from the pK_a values of the acids by calculating an equilibrium constant for the reaction.

For example, in this reaction, the pK_a of the stronger acid, when subtracted from the pK_a of the weaker acid, gives the logarithm of the equilibrium constant. The antilog of 2 gives an equilibrium constant of 10².

The large value of the equilibrium constant indicates that the equilibrium lies to the side of the weaker, more stable acid having a higher pK_a value.