Acid-Base Equilibria (6) Buffer solutions and Buffer calculations

Edexcel A level Chemistry (2017)

Topic 12: Acid–Base Equilibria:

Here are the next learning objectives:

12/18. To know what is meant by the term ‘buffer solution’

12/19. To understand the action of a buffer solution

12/20. To be able to calculate the pH of a buffer solution given appropriate data

12/21. To be able to calculate the concentrations of solutions required to prepare a buffer solution of a given pH

Buffer Solutions and Buffer calculations

A buffer solution changes pH slightly or very little (the point to make is that the pH does NOT remain constant but changes little) when a strong acid solution or a strongly alkaline solution is added to it.

A buffer solution tends to resist change in pH.

However if sufficient strong acid solution or strongly alkaline solution is added the buffer breaks down completely.

A typical acid buffer solution is formed from equal amounts of a weak and its weak acid salt. 

So a solution that is 0.1M with respect to ethanoic acid and 0.1M with respect to sodium ethanoate is an acidic buffer solution. 

Buffer Action

Buffer solutions work like this:

Addition of a small volume of strong acid contains H+ ions.

These H+ ions have an effect on this equilibrium:

CH₃COOH   +   H₂O    ⇌  CH₃COO^—  +   H₃O+

Rearranging gives

So if  [H₃O+ ] is to remain constant the ratio of undissociated acid anion to undissociated acid has also to remain constant.

The addition of the sodium ethanoate to the ethanoic acid ensures there is an excess of ethanoate anions and these push the equilibrium to the left reducing the H+ ion concentration and giving the buffer solution a higher acid pH than the pH of ethanoic acid.

Any additional H+ ions added will react with the ethanoate anions and form more ethanoic acid pushing the equilibrium to the left, slightly reducing the pH.

Similarly if a small amount of strong alkali is added the OH^— ions with combine with the H+ ions to form water and the equilibrium will move slightly to the right increasing the pH slightly to compensate.

The arrows indicate the direction the position of equilibrium moves in the buffer with addition of H+ or OH^—  ions.

As you must realise, the changes above do affect the ratio of undissociated acid anion to the undissociated acid and hence the pH.

Buffer calculations

Now if a buffer solution contains 0.1M ethanoic acid and 0.1M sodium ethanoate then we can calculate its buffer pH.

We must assume two things:

a)  that we can ignore the dissociation of water to form H3O+ ions

b)  that the weak acid ethanoic acid is so poorly dissociated that the dissociation does not affect the value of the concentration of the acid or that of the acid anion. 

Therefore use this form of the K_a equation:

Substituting the values for each species on the right hand side gives

[H₃O+] =  1.8  ×  10^—5

pH =   4.74

Now suppose we want to prepare a buffer solution of pH 5.7 from ethanoic acid and sodium ethanoate then again we use this equation:

we know that the pH of the buffer is 5.7 and therefore the [H₃O+] is 10^—5.7

so    [H₃O+]   =  1.995  ×  10^—6 mol.dm^—3

if                                        Ka = 1.8  ×  10^—5   mol.dm^—3

then

then

therefore the ratio of [CH₃COOH]  :  [CH₃COO^—] is 1:11

If the volume of the buffer to be made up is exactly 1 dm³

then the litre should contain 11moles sodium ethanoate for every 1 mole of ethanoic acid from which the exact mass or liquid quantities can be determined. 

You should now be able to calculate the pH of buffers given their composition and the composition of a buffer given the pH required. 

Examples:

What is the pH of a buffer solution containing 0.05M ethanoic acid and 0.2M sodium ethanoate?

Given the pH required is 6.1 how can ethanoic acid and sodium ethanoate be sued to prepare this buffer?