IB syllabus > acids & bases (hl) > 18.2 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

18.2 - Buffer Solutions

18.2.1: Describe the composition of a bufer solution and explain its action.

What is a buffer solution?

Some solutions resist changes in pH when small amounts of acid or base are added. On addition of acid the hydrogen ions get removed by one of the components of the mixture and on addition of base the hydroxide ions get removed by one of the components of the mixture. The effect is called buffering action an solutions that behave this way are called buffers.

It must be remembered at this point that pH is a measure of the concentration of H+ ions. Also that the product of H+ ion and OH- ion concentrations is always constant (at constant temperature) and equal to 1 x 10-14

[H+] x [OH-] = 1 x 10-14 mol2 dm-6

There are two types of buffer.

  1. Weak acid and the salt of the same weak acid, (for example a solution containing ethanoic acid and sodium ethanoate). This gives a buffer solution with a pH less than 7
  2. Weak base and the salt of a the same weak base (for example ammonia and ammonium chloride solution). This gives a buffer with a pH greater than 7

The first (acidic) buffer works in the following way.

If an acid is added it combines its free hydrogen ions with the ions from the salt of the weak acid making molecular weak acid that cannot affect the pH.

If a base is added the OH- ions from the base react with the H+ ions that are present from the weak acid dissociation. Having been removed from the solution this stimulates the weak acid to produce more H+ ions (Le Chatelier's Principle) and the original pH is re-established.

Example - an acidic buffer:

A mixture of ethanoic acid and sodium ethanoate gives a buffer with a pH in the acidic range (there are free hydrogen ions in the solution).

Two equilibria are established

equilibrium 1: CH3COOH CH3COO- + H+

The first equilibrium lies 99% to the left hand side (i.e. there is a large store of ethanoic acid molecules)

equilibrium 2: CH3COONa CH3COO- + Na+

The second equilibrium lies almost 100% to the right hand side (i.e. there is a large store of ethanoate ions)

Hence, the mixture has large quantities of both ethanoic acid molecules [CH3COOH] from the first equilibrium, and ethanoate ions [CH3COO-] from the second equilibrium.

Addition of small quantities of acid (H+)

The H+ ions added react with the excess ethanoate ions in equation 2 and are removed from the solution as ethanoic acid molecules (these have no effect on the pH). Hence the pH stays the same.

Addition of small quantities of base (OH-)

In this case, the OH- ions react with the hydrogen ions from the first equilibrium removing them from the right hand side. There is, however, a large reservoir of ethanoic acid on the left hand side of this equilibrium able to dissociate and make more hydrogen ions, restoring the pH.

Example: A basic buffer

A mixture of ammonium chloride (salt of a weak base) and ammonia solution (weak base) has buffering action.

Once again two equilibria are established:

equilibrium 1: NH3 + H2O NH4+ + OH-

This equilibrium lies very much to the left hand side (i.e. there is a large reserve of free ammonia molecules)

equilibrium 2: NH4Cl NH4+ + Cl-

This equilibrium lies 100% to the right hand side (i.e. there is a large reserve of ammonium ions)

Addition of small quantities of acid (H+)

The H+ ions from the acid react with the OH- ions from equilibrium 1 and remove them. However, there is a large reserve of ammonia molecules available to dissociate from the left hand side making more OH- ions restoring the pH (remember that the value of [H+][OH-] must be constant under all conditions except a change of temperature)

Addition of small quantities of base (OH-)

Adding more OH- ions that can react with the free ammonium ions (from equilibrium 2) producing more ammonia (as in equilibrium 1) and effectively being removed from the system. The ammonia molecules have no effect on pH an therefore the pH remains the same.

18.2.2: Solve problems involving the composition and pH of a specified buffer system. Only examples involving the transfer of one proton will be assessed. Examples should include ammonia solution/ammonium chloride and ethanoic acid/sodium ethanoate. Students should state when approximations are used in equilibrium calculations.. Use of quadratic equations will not be assessed. AIM 7: Virtual experiments can be used to demonstrate this.

The simplest way of preparing a buffer solution is to dissolve a known quantity of the salt of the weak acid (or base) in a solution of weak acid (or base ) of known concentration.

Example: Sodium ethanoate is the salt of a weak acid - it has the formula CH3COONa (relative molecular mass = 82)

If 8.2 g of sodium ethanoate are dissolved in 100 cm3 of ethanoic acid then:

The concentration of the sodium ethanoate is equal to 0,1/0,1 = 1M

The concentration of the ethanoic acid = 1M

Therefore the buffer solution will have pKa = pH

Therefore the buffer has a pH value of 4,78

Another way to prepare a buffer solution (much favoured by the IB examiners) is to neutralise an excess of weak acid (or weak base) with some strong base (or strong acid). The neutralisation produces the salt of the weak acid (or base) 'in situ' and, as the weak acid was in excess, there will still be some weak acid in the mixture. The resultant mixture contains both the salt of the weak acid and the weak acid itself.

Example: Add 50cm3 sodium hydroxide 1M solution to 50 cm3 1M ethanoic acid

Moles of sodium hydroxide added = 0,05 x 1 = 0,05 moles

This will neutralise and equal number of moles of the ethanoic acid = 0,05 moles and produce exactly 0,05 moles of sodium ethanoate


The concentration of the sodium ethanoate produced:

= 0,05 moles / 0,1 dm3 = 0,5M

The total initial acid moles :

= 1 x 0,1 = 0,1 moles

Moles of acid reacted with sodium hydroxide = 0,05 moles

Therefore moles of acid remaining

= 0,1 - 0,05 = 0,05 moles

molarity of acid = 0,05 /0,1

molarity of acid = 0.5 M

The buffer solution pH can then be obtained form the buffer law

pKa = pH + log[ethanoate ion]/[ethanoic acid]

pH = pKa - log[ethanoate ion]/[ethanoic acid]

pH = pKa - log 1

pH = pKa

In this case it can be easily appreciated that when the concentration of the acid and the salt are equal then pH =pKa

This represents a rapid shortcut in questions of this type. If it can be shown that the weak acid concentration (or the weak base concentration) equals the salt concentration then pH = pKa and NO calculation needs to be carried out