What is a buffer and how do I calculate its pH?

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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

  1. 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.
  1. 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.
  1. 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.
  1. 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?
  2. Calculate the pH of a buffer made from adding 9.08g of sodium ethanoate, CH3COONa, to 200 cm3 of 0.80 mol dm-3 ethanoic acid (Ka = 1.75 x 10-5 mol dm-3).
  3. What mass of sodium methanoate, HCOONa, must be added to 250 cm3 of 0.5 mol dm-3 methanoic acid to make a buffer solution of pH 4.1, given that the Ka of methanoic acid is 1.58 x 10-4 mol dm-3?
  4. A buffer solution contains 0.025 mol of sodium ethanoate dissolved in 500 cm3 of 0.07 mol dm-3 ethanoic acid (Ka = 1.76 x 10-5 mol dm-3). A sample of 5 cm3 of 2.0 mol dm-3 HCl is added to this buffer. Calculate the pH of the solution formed. 
  5. 300 cm3 of buffer solution was prepared from 0.260 mol dm-3 ethanoic acid (Ka = 1.74 x 10-5 mol dm-3) and 0.120 mol dm-3 sodium ethanoate. 4.20 x 10-3 mol of sodium hydroxide was added to the buffer solution. Calculate the pH of the resulting solution. 
  6. (a)  When sodium ethanoate is added to food as an acidity regulator, a buffer consisting of sodium ethanoate and ethanoic acid is formed. Explain, with the help of an equation, how this buffer works when acid is added.

(b)  Calculate the pH of a buffer containing ethanoic acid (Ka = 1.7 x 10-5 mol dm-3) and sodium ethanoate in a 2:1 ratio.

(c)  Some students set out to make a buffer of pH 5.50 using 35.0 cm3 of 0.100 mol dm-3 ethanoic acid. What mass of sodium ethanoate must they add to the acid to make their buffer?

10. Benzoic acid, C6H5COOH, is a weak monoprotic acid.

(a)  Write the expression for Ka for benzoic acid. 

(b)  A buffer was made from benzoic acid and its salt sodium benzoate, C6H5COONa, using the following method:

  • 3.0g of sodium benzoate was dissolved in 100 cm3 of distilled water
  • 150.0 cm3 of 0.10 mol dm-3 HCl was added to the solution
  • the volume was made up to 600 cm3 with distilled water

(i) Write an equation for the reaction of aqueous sodium benzoate with HCl.

(ii) Calculate the pH of the buffer solution (the pKa of benzoic acid is 4.2).

11. Fumaric acid acts an acid, HA, in aqueous solution.

(a) Write the equilibrium for the ionisation of a weak acid, HA, in solution.

(b) Calculate the pH of a 0.14 mol dm-3 solution of fumaric acid (Ka = 9.3 x 10-4 mol dm-3).

(c) Describe the approximations you made in carrying out the calculation in part (b). Explain which approximation is likely to lead to the greater inaccuracy in your answer.

(d) A mixture of fumaric acid and sodium fumarate can be used as a buffer solution to regulate the acidity levels in food. Describe what is meant by a buffer solution and explain how a buffer system works based on the equilibrium in part (a).

(e) Calculate the pH of a fumaric acid / sodium fumarate buffer solution where the concentration A– is equals that of HA.

Answers

9. (a) CH3COOH (aq)  ⇌  CH3COO–(aq)  +  H+(aq)   .  
A buffer solution is able to resist changes in pH when small amounts of acid or alkali are added – it is an equilibrium system made from a weak acid (ethanoic acid) and its conjugate base (sodium ethanoate). When small amounts of acid are added, the H+ ions react with the ethanoate ions present (the ethanoate ions act as a sink for the extra H+ ions ), shifting the equilibrium position to the left. As the [H+] remains constant, there is no change in pH.


10.

11.

(c) Assumption 1:  HA dissociates to give H+ and A– in a 1:1 ratio, so [H+]=[A–].

Assumption 2: the concentration of HA at equilibrium is the same as the initial concentration of HA i.e. as it is a weak acid we assume that a negligible amount of HA has actually dissociated.

(d) A buffer solution is able to maintain its pH when small amounts of acid or alkali are added.  Addition of acid / H+ ions causes equilibrium position to shift to the left as the high concentration of A– ions from the salt react with the H+ ions to form more HA. A– acts as a sink for the additional H+ ions.

Addition of alkali / OH– ions causes the equilibrium to shift to the right. HA dissociates to produce H+ ions which react with the added OH– ions. HA acts as a source or reservoir of extra H+ ions. In either case, changes to the pH are minimised.Â