Weak Acid Equilibria

We are often interested in solving weak acid (base) equilbria problems. While there are a huge number of such potential applications, there is one type of problem in particular that comes up over and over again. These problems give some initial concentration of acid (base) and seek the concentrations of chemical species at equilibrium. In particular, for acid problems we are interested in the hydronium ion concentration, [H₃O⁺]. For base problems, we are usually interested in the hydroxide ion concentration, [OH^-].

Let's examine a typical weak acid problem. You'd like to know the hydronium ion concentration, [H₃O⁺] for a 0.1 M solution of benzoic acid. Benzoic acid is a weak acid (K_a= 6.4 x 10^-5) so you know that the concentration will be very low. This is an equilibrium problem so you'll want to look at both the chemical equation and the equilibrium constant. We write the same equation for every weak acid in water. The acid plus the water react to form the hydronium ion and the deprotonated acid (the benzoate ion).

\[\rm{C_6H_5COOH(aq) + H_2O(l) \rightleftharpoons C_6H_5COO^-(aq) + H_3O^+(aq)}\]

The equilibrium constant for this reaction is

\[{\rm K_a = {{[H_3O^+][C_6H_5COO^-]} \over {[C_6H_5COOH]}}}\]

We can find the [H₃O⁺] at equilibrium if we know the initial concentration of benzoic acid and the K_a for benzoic acid. The initial concentration of the acid is 0.1 M. K_a= 6.4 x 10^-5.

To solve this problem we can set up a RICE table. We know the initial concentration of benzoic acid is 0.1M. We will assume the concentrations of the other ions are zero (this is subtle approximation for the hydronium ion). We will choose as our variable "x" the concentration of the acid that dissociates.

R	C6H5COOH	H2O	C6H5COO-	H3O+
I	0.1	-	0	0
C	-x	-	+x	+x
E	0.1-x		+x	+x

Putting the equilibrium values into the mass action expression we can solve for the amount of acid that dissociated.

\[{\rm K_a = {{[H_3O^+][C_6H_5COO^-]} \over {[C_6H_5COOH]}}}= {{(x)(x)} \over (0.1-x)}\]

Since we know K_a this can be re-arranged to yield a quadratic equation that we can solve. However, we can very easily arrive at an approximate answer. This is because we know that "x" will be small. That is that it is a small number compared to the initial concentration. How do we know this? Benzoic acid is a weak acid. This means the equilibrium favors the reactants. K_a is small. As a result we know what the concentration of the acid will be at equilibrium. It will be essentially unchanged. As a result we can approximate the equilibrium concentration of (0.1-x) ≈ 0.1. Putting this and the value of K_a for benzoic acid into the equation we find.

\[{\rm K_a }= 6.4 x 10^{-5}= {{(x)(x)} \over (0.1-x)} \approx {{(x)(x)} \over (0.1)}= {x^2 \over 0.1}\]

This is simple to solve and we find

\[ x = \sqrt{(6.4 x 10^{-5})(0.1)} = 2.5x10^{-3}\]

Now it is important to remember how we defined "x". In this case it was the concentration change for the acid. It also happens that for this situation, the hydronium ion concentration [H₃O⁺]=x.

This problem can be solved in the exact same manner for any weak acid in some concentration in pure water. If we call the weak acid HA with an initial concentration, C_HA we can find the concentration of all the species at equilibrium if we know K_a for the acid.

\[\rm{HA(aq) + H_2O(l) \rightleftharpoons A^-(aq) + H_3O^+(aq)}\]

The equilibrium constant for this reaction is

\[{\rm K_a = {{[H_3O^+][A^-]} \over {[HA]}}}\]

Setting up the RICE table for an initial concentration of C_HA for the weak acid HA. And defining the number of moles of the acid that dissociates as "x" yields.

R	HA	H2O	A-	H3O+
I	Ca	-	0	0
C	-x	-	+x	+x
E	Ca-x		+x	+x

Putting this into the equilibrium expression we find

\[{\rm K_a = {{[H_3O^+][A^-]} \over {[HA]}}}= {{(x)(x)} \over (C_{HA}-x)}\]

We can again make the assumption that the change in concentration of the acid is negligible. This gives

\[{\rm K_a }= {{(x)(x)} \over (C_{HA}-x)} \approx {{(x)(x)} \over (C_{HA})}= {x^2 \over C_{HA}}\]

Solving this we get

\[ x = \sqrt{{\rm (K_a)(C_{HA})}}\]

Remembering that we defined "x" as the change in concentration of the weak acid we know that this is also the concentration of the hydronium ion at equilibrium. This yields a very useful approximate formula, that the concentration of hydronium ion at equilibrium in a solution that contains only a weak acid in water is given by

\[ {\rm[H_3O^+] = \sqrt{(K_a)(C_{HA})}}\]

We can arrive at a nearly identical result relating the concentration of a weak base and K_b to the hydroxide ion concentration [OH^-]. The chemical equation for a weak base, B, in water is

\[\rm{B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq)}\]

The equilibrium constant is

\[{\rm K_b = {{[OH^-][BH^+]} \over {[B]}}}\]

Using an identical derivation, we arrive at the approximate result where C_B is the concentration of the weak base.

\[ {\rm[OH^-] = \sqrt{(K_b)(C_B)}}\]

When do these approximations fail? This question is difficult to answer, since it all depends on how how accurate an answer you are trying to find. They are always approximate. However, the approximation is better or worse in certain limits. This can be understood by examining what is being assumed in the approximation. There are two key assumptions. First, it is assumed that the initial concentration of hydronium ion (hydroxide ion) in pure water is zero. This is wrong. The concentration in pure water is very very small, but it is not zero. As a result, this approximation will fail when the solution is very dilute or the acid (base) is so weak that it essentially does not dissociate. The second assumption is that the amount of the acid that dissociates is so small that the concentration is essentially unchanged [ C_HA - x ≈ = C_HA ] When is this a reasonable approximation? When C_HA >> x. This will be the case when the solution is concentrated and the acid is weak. Thus this approximation is better for concentrated solutions of weak acids with small K_a's.