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I have previously made two blog posts on acids and bases. The idea of these posts is to either give you a heads up on some content that you have still to cover or to revise some concepts you may already have covered.

This post will be the third and final one on acids and bases and I thought I would use this one to cover the pH of weak acids. This would be a HL concept.

In order to understand weak acids, you will have already needed to cover (and understand) the equilibrium topic, so it may be worth getting these notes to hand as well.

As a rule of thumb, a weak acid will be an organic acid. It will be made of carbon, hydrogen and oxygen and will have the COOH functional group.

Weak acids are different to strong acids because in solution they will only partially dissociate. Most of the solution of weak acid will stay as the weak acid, and a small amount will dissociate into the anion and hydrogen ion.

For example, using ethanoic acid as an example:

CH3COOH ⇌ CH3COO +H+

Note that because there are ‘products’ and ‘reactants’ we use the equilibrium symbol to depict this reaction.

As we have an equilibrium, it reasons that we could write an expression for Kc … and we do, but this time we call it Ka (‘a’ for acid) – it is also possible to get Kb for weak bases.

The crucial point to understand here is that we have a constant – this will be needed to calculate the pH.

The expression for the equilibrium could be shown:

Ka = [CH3COO] x [H+] / [CH3COOH]

If you are going to calculate the pH of this weak acid, you will need the Ka value. This can be found in the data booklet … or can it? You will see in Table 21 of the data booklet (Strengths of organic acids and bases) that a pKa value is shown … not a Ka value.

However, the concept of pKa and Ka is exactly the same as [H+] and pH. The ‘p’ means –log10 and it puts a range of very small numbers into more meaningful values.

So to turn a Ka into pKa carry out the operation:

pKa = -log10 (Ka)

Which also means

10-pKa = Ka

To calculate the pH of the weak acid you will also need to use a little trick. The concentration of anion and H+ will be the same, as they are found in a 1:1 ratio.

This means that our expression:

Ka = [CH3COO] x [H+] / [CH3COOH]

Can be re-written:

Ka = [H+]2 / [CH3COOH]

A simple rearrangement gives us:

[H+]2 = Ka x [CH3COOH]

And therefore:

[H+] = √ Ka x [CH3COOH]

Once you have determined [H+], don’t forget to work out the pH by using –log 10 [H+]. It is amazing how many people forget this last step!!

I have not covered all of the concepts in this unit but the ones covered will certainly help you with your studies.