A buffer is defined as a solution that is able to resist changes in pH when small amounts of acid or alkali are added. It is made from either a weak acid and its salt (conjugate base) or from a weak base and its salt (conjugate acid). By choosing suitable conjugate acid-base pairs and adjusting their relative concentrations, we can make a buffer solution with almost any pH.
Consider a buffer made from 0.1 mol dm-3 ethanoic acid and 0.1 mol dm-3 sodium ethanoate. The action of the buffer depends on the equilibrium:
The buffering action of this solution can be explained using Le Chatelier’s Principle.
Addition of acid / H+ ions: they will react with some of the ethanoate ions, A–, forming more ethanoic acid, HA, and the equilibrium position shifts to the left. The overall [H+] is maintained so there is no change in pH. We can say that [A–] acts as a SINK for the extra H+ ions that have been added.
Addition of alkali / OH– ions: they will react with the H+ ions in the buffer (forming water) which causes the equilibrium position to shift to the right. More ethanoic acid molecules, HA, will dissociate to replace the ‘used up’ H+ ions and so any change in pH is minimised. We can say that [HA] is a SOURCE or RESERVOIR of extra H+ ions.
Calculating the pH of a buffer
Let’s go back to our general expression for a buffer solution, remembering that HA is the weak acid and A– is the salt / conjugate base used to make this buffer.
HA (aq) ⇌ H+ (aq) + A– (aq)
Alternatively, we could take logarithms of both sides of the equilibrium expression:
In practice, the buffer is most effective at a pH that is equivalent to the pKa of the weak acid used (or at least within +/- 1 unit). This is sometimes described as the coarse-tuning of the buffer.
For example, if I want a buffer of pH 5, then choosing ethanoic acid as the weak acid is a good idea as it has a pKa of 4.8.
An efficient buffer solution should be equally resistant to the addition of acid or alkali, and so would need to have an equal concentration of A– and HA. We can fine-tune the buffer by altering this ratio, but moving past a 1:3 / 3:1 ratio of concentrations means we are in danger of not having enough HA or A– to buffer effectively.
Practice questions
- Calculate the pH of a buffer solution made from 0.50 mol dm-3 methanoic acid (Ka = 1.6 x 10-4 mol dm-3) and 2.5 mol dm-3 potassium methanoate.
- Calculate the pH of a buffer solution containing 2.0 mol dm-3 butanoic acid (Ka= 1.5 x 10-5 mol dm-3) and 1.0 mol dm-3 potassium butanoate.
- Determine the ratio of 0.5 mol dm-3 ethanoic acid : 0.5 mol dm-3 sodium ethanoate needed to make a buffer solution of pH 4.40, given that the pKa of ethanoic acid is 4.74.
- What amount of sodium methanoate must be added per dm3 of 0.55 mol dm-3 methanoic acid to make a buffer solution of pH 3.8, given that the Ka of methanoic acid is 1.58 x 10-4 mol dm-3?
Answers
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