Equilibrium Calculations

Introduction

We have seen that there is a relationship between the pK_a values of the conjugate acids involved in an acid-base reaction and the value of K_eq for that reaction. Since the principles involved in calculating K_eq for an acid-base reaction may be extended to other reactions, we will take the time to derive the relationship between pK_a values and K_eq.

Simultaneous Equilibria

Consider the reaction between acetic acid and trifluoroacetate ion shown in Equation 1.

The expression for the equilibrium constant for this reaction is

but how do you determine the numerical value of K_eq? A simple way involves treating each acid and its conjugate base separately. Consider the dissociation of acetic acid in water as shown in Equation 2.

As we have already seen, the equilibrium expression for this reaction is

Equation 3 describes the dissociation of trifluoroacetic acid in water.

The equilibrium expression for this reaction is

Dividing the equilibrium expression for Equation 2 by that for Equation 3 gives

Clearing terms and rearranging yields

which is the equilibrium expression for Equation 1. So, by independently measuring the degree of ionization of each acid in water, we can calculate the equilibrium constant for a reaction involving the two acid/conjugate-base pairs.

It is possible to arrive at the value of K_eq for Equation 1 by another method. The pK_a of acetic acid is 4.75, while that of trifluoroacetic acid is 0.18. The difference in these values is (0.18-4.75) = -4.57. The antilogarithm of -4.57 = 2.69 x 10^-5. Fortunately, if you use integral values for the pK_a values of each acid, you can do the math in your head. Using integral pK_a values for acetic acid (5) and trifluoroacetic acid (0) gives a difference of (0-5) = -5, and the antilogarithm of -5 = 1 x 10^-5. Conversely, the antilogarithm of a positive integer, say 5, is 1 x 10⁵.

Doing It In Your Head

In order to calculate the equilibrium constant for any acid-base reaction follow these steps:

1. Write out the equation for the reaction.
2. Identify the acid of each conjugate acid-base pair
3. Assign the correct pK_a value to each acid.
4. Subtract the pK_a value of the reactant acid from the pK_a value of the product acid. Call the difference D.
5. K_eq = 1 x 10^D

Here's an example using the reaction of phenol with sodium hydroxide to produce the phenolate ion and water.

1. Write out the equation for the reaction.
2. Identify the acid of each conjugate acid-base pair.
3. Assign the correct pK_a value to each acid.
4. Subtract the pK_a value of the reactant acid from the pK_a value of the product acid. Call the difference D. D = (16-10) = 6
5. K_eq = 1 x 10^D
6. K_eq = 1 x 10^D K_eq = 1 x 10⁶

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Extensions to Other Systems

The value of this type of calculation lies in its application to related systems. Consider the simple substitution reaction shown in Equation 4.

This is not an acid-base reaction in the same sense that the previous examples were. There is no proton transfer from one atom to another. Still, we can assess the equilibrium constant by considering the relative stabilities of the hydroxide and chloride ions. If chloride ion is less basic than hydroxide ion, the equilibrium will favor the products. If hydroxide ion is less basic than chloride ion, the equilibrium will favor the reactants. To assess the relative stabilities of chloride and hydroxide ion we need to write an equation for a reaction involving these two ions and their conjugate acids. Equation 5 is one possibility.

Using the method outlined above, it is easy to determine that the equilibrium constant for this reaction is 1 x 10²³. In other words, chloride ion is much less basic than hydroxide ion. As a first approximation, the difference in stability of chloride and hydroxide ion should be the same in Equation 4 as it is in Equation 5. Thus it is reasonable to conclude that the equilibrium constant for reaction 4 should also be approximately 1 x 10²³.

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