Lesson 7: Equilibrium Constants (K_a & K_b)

Values of K_a range from 10¹² all the way to 10^-50, an enormous range. Most common acids, however, have K_a values in the range of 10^-12 to 10⁶. A strong acid generally is one in which K_a > 1. A useful classification says that strong acids are stronger than the H₃O⁺ ion (K_a = 1), weak acids are weaker than the H₃O⁺ ion but stronger than H₂O (K_a = 10^-14), while very weak acids are weaker acids than H₂O.

The acetate ion is the conjugate base of acetic acid, a weak acid, thus the acetate ion is a weak base. The reaction is shown with arrows going in both directions.

The equilibrium constant for this reaction is given the symbol K_b, and is often called the base constant. Like K_a, K_b will be large for a strong base and small for a weak base.

Again, water is left out of the equilibrium constant expression because it is the solvent. Its concentration does not change appreciably as the reaction proceeds.

Also, as with the K_a, the equilibrium constant expression is written by multiplying the product concentrations and dividing by the reactant concentrations.

a.

b.

c.

You can use the equilibrium constant expression to calculate the concentration of the hydronium ion in solution. Here is example 9a taken from your workbook.

The equilibrium constant expression can be used in other ways as well, as we’ll see in the next few examples.

For this problem, we want the pH. Recall that:

pH = -log[H⁺]

Therefore we will need to calculate [H⁺] first, then find its negative log to get the pH.

First we solve the equilibrium constant expression for [H+]. Then we substitute the values for the known concentrations and the Ka. Finally, we take the negative log of the result.

In a logarithm, the significant digits are those that follow the decimal point. Since 1.8 x 10-5 has two significant digits, we keep two digits after the decimal point in the pH.

Example 9b asks for the acetate and hydronium ion concentrations and the pH of a 0.2 M solution of acetic acid.

The pH can be calculated once we know the value of [H₃O⁺], since pH is just the negative log of the hydronium ion concentration. To determine the concentration of the acetate ion we will make use of the fact that when the acid dissociates, it releases one acetate ion for each hydronium ion, thus the concentration of these two ions must be the same. We know the value of the Ka from the previous example.

The concentration of acetic acid given, 0.2 M, refers to the concentration of the acid before it dissociates. Since this is a weak acid, we will start by assuming that only a very small fraction of the acid molecules dissociate so that, at equilibrium, the concentration of the acid is still very nearly 0.2 M.

Acid concentrations are commonly given as the number of moles of acid in solution before any protons have been donated. Once the reaction has reached equilibrium, the amount of undissociated acid will depend on how strong the acid is. For a strong acid, this will be essentially zero. However, for weak acids, only a tiny fraction of the acid dissociates, so that the concentration of undissociated acid at equilibrium will be very nearly equal to its original value.

Since, when the acid dissociates, one acetate ion will be left behind each time one proton is donated, the concentration of acetate ions and the concentration of H₃O⁺ ions will be equal. We’ll call the concentration x.

In water there is a very small number of H₃O⁺ ions present from the water itself. This is too small to matter unless the acid is very weak, that is, its K_a is less than about 10^-14. Since the K_a for acetic acid is 1.8 x 10^-5, we can ignore the small concentration of H⁺ ions from the water and our assumption that the number of hydronium ions is the same as the number of acetate ions is valid.

From the previous example, we know that K_a = 1.8 x 10^-5. We can now substitute these values into the equilibrium constant expression and solve for x. This tells us both the acetate and hydronium ion concentrations.

Since 0.0019 mol/liter of H⁺ ions are present, this same amount of acid must have dissociated, so that the actual concentration of acid is just 0.2- 0.0019, or 0.1981 M, not 0.2 M. This represents an error of less than 1%.

Since we have the concentration to only one significant digit (0.2 M), an uncertainty of about 10%, our assumption is justified.

To calculate the pH, we simply take the negative log of the [H⁺]. The pH of this solution is 2.7.

If you would like to see how to calculate the [H⁺] more accurately, taking into account both the dissociation of the acid and the presence of a small amount of H⁺ from water (and you are comfortable solving quadratic equations), ask the instructor when you visit the lab.

To calculate the value of the K_a, we will need the values of the concentration of lactic acid, lactate ion, and H⁺ at equilibrium. We will essentially work the previous problem in reverse, in the process making the same set of assumptions.

We begin by assuming, as we did in the previous example, that the concentration of lactic acid is the same before and after equilibrium is established, 0.02 M.

If all the lactic acid dissociated, the [H⁺] would be 0.02 M. The negative log of this number is 1.7. Since the pH of this solution is 2.8, we can assume we are dealing with a weak acid and that our assumption is reasonable.

We can determine the [H⁺] from the pH. We can also assume, as before, that the concentration of lactate ions and the concentration of H⁺ ions will be the same, since each time a lactic acid molecule dissociates, one of the each ion is released.

Again we ignore the small difference in these concentrations due to H⁺ ions from water. In general, the concentration of H+ ions from water will be less than 10^-7 M, a good deal smaller than the 1.6 x 10^-3 moles per liter of H⁺ from the acid.

Now we substitute these values into the equilibrium constant expression to calculate K_a.

Strictly speaking, we should round this answer down to 1 x 10^-4 since the concentration of lactic acid was given to only one significant digit. Also, if you paid attention to the units in this problem, you noticed that the units of the answer are M, moles per liter, buth that we didn’t include them in the answer.