5.4:
Position of Equilibrium in Acid-Base Reactions
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 pKa values of each acid.
Here, ethanol with a pKa of 15.9 is a stronger acid than ammonia, whose pKa 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 pKa values of the acids by calculating an equilibrium constant for the reaction.
For example, in this reaction, the pKa of the stronger acid, when subtracted from the pKa 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 pKa value.
5.4:
Position of Equilibrium in Acid-Base Reactions
In any solution, the value of pKa indicates whether an acid is completely dissociated or not. A negative pKa corresponds to a stronger acid, whereas a positive pKa corresponds to a weaker acid. Consider the reaction between ammonia and an ethoxide ion. In this reaction, ethanol with a pKa of 15.9 is a stronger acid than ammonia with a pKa of 38. Recall that the strong acid forms a weak conjugate base, and a weak acid forms a strong conjugate base. Hence, the ethoxide ion is a weak base.
Figure 1. Acid–base reaction between ammonia and an ethoxide ion
In an equilibrium-controlled acid–base reaction, the equilibrium position always favours the formation of the weaker acid and the weaker base. This is because the weaker acid and the weaker base are the most stable species due to their lower potential energies. Therefore, in this reaction, the equilibrium favours the formation of the ethoxide ion and ammonia.
Additionally, the position of the equilibrium can also be identified using the equilibrium constant that is calculated from pKa values. For an acid–base reaction, the equilibrium constant, Keq, can be calculated by subtracting the pKa value of the acid on the left side from the pKa value of the acid on the right side and then taking the antilog of the result. Depending on the value of the equilibrium constant, the position of the equilibrium is determined. When the value of Keq is greater than 1, the position of the equilibrium lies far to the product side. On the other hand, with the equilibrium constant smaller than 1, the equilibrium lies far to the reactant side.
Suggested Reading
- Brown, W.H., & Iverson, B.L., & Anslyn, V.E., & Foote S.C. (2014). Organic Chemistry. Mason, Ohio: Cengage Learning, 118-123.
- Solomons, G., & Fryhle, C. & Snyder, S. (2015). Organic Chemistry. New Jersey, NJ: Wiley, 192-194.
- Loudon, M., & Parise, J. (2016). Organic Chemistry. New York, NY: Macmillan Publishers, 230-234.
- Klein, D. (2017). Organic Chemistry. New Jersey, NJ: Wiley, 183-188.