### Using pK_{a} values to predict reaction equilibria

By definition, the pK_{a} value tells us the extent to which an acid will react with water as the base, but by extension, we can also calculate the equilibrium constant for a reaction between any acid-base pair. Mathematically, it can be shown that:

K_{eq} (for the acid base reaction in question) = 10^{Δ}^{pKa}

where **Δ**pK_{a} = pK_{a} of product acid minus pK_{a} of reactant acid

Consider a reaction between methylamine and acetic acid:

First, we need to identify the acid species on either side of the equation. On the left side, the acid is of course acetic acid, while on the right side the acid is methyl ammonium. The specific pK_{a} values for these acids are not on our very generalized pK_{a} table, but are given in the figure above. Without performing any calculations, you should be able to see that this equilibrium lies far to the right-hand side: acetic acid has a lower pK_{a}, is a stronger acid, and thus it wants to give up its proton more than methyl ammonium does. Doing the math, we see that

K_{eq} = 10^{Δ}^{pKa} = 10^{(10.6 – 4.8)} = 10^{5.8} = 6.3 x 10^{5}

So K_{eq}is a very large number (much greater than 1) and the equilibrium lies far to the right-hand side of the equation, just as we had predicted. If you had just wanted to approximate an answer without bothering to look for a calculator, you could have noted that the difference in pK_{a} values is approximately 6, so the equilibrium constant should be somewhere in the order of 10^{6}, or one million. Using the pK_{a} table in this way, and making functional group-based pK_{a} approximations for molecules for which we don’t have exact values, we can easily estimate the extent to which a given acid-base reaction will proceed.

### Example

__Exercise 7.3__ Show the products of the following acid-base reactions, and estimate the value of K_{eq}.

### Contributors

**Organic Chemistry With a Biological Emphasis**by Tim Soderberg (University of Minnesota, Morris)