Titration calculations

There are a variety of possible calculations that you may be set. I shall show you how to work out the most common calculation - that is working out the unknown concentration of an acid or alkali. There are quite a number of steps, but it does get easier with a real example and practice. It also helps if you label your calculation steps carefully.

Step 1 - Average titre

If you have more than one titre value you will have to decide on what value to use. Generally this will involve averaging the values. However, do not include any values that are clearly inaccurate. Remember, you should have continued your titration work until you obtained concordant results. These agree to within 0.1 cm³. For example, if you had three titres, 12.55 cm³, 12.10 cm³, and 12.20 cm³, ignore the first value as it is too far out, and average the other two. This gives you an average titre of 12.15 cm³. The values have been given to 2 decimal places to show that the burette has been read to the nearest 0.05 cm³.

Step 2 - Number of moles of titrant

The titrant is the solution you titrated with. This is in the burette and you know its concentration accurately. You also know the volume used - it is the average titre that we have just worked out. Whenever you see a calculation in which you know the volume and concentration of a substance you can (and should!) work out the number of moles. This can be done with the correct moles equation. Don't forget to convert cm³ into dm³ and label the line correctly. For example,

Number of moles of HCl = 0.100 × 12.15/1000

Step 3 - Write down the equation and reacting ratio

We need to write down a balanced chemical equation for the reaction between the acid and the alkali. This then allows us to look at the link between the number of moles of acid and the number of moles of alkali. For example, in the reaction between hydrochloric acid and sodium hydroxide solution the equation is:

NaOH(aq) + HCl(aq) ⇒ NaCl(aq) + H₂O(l)

The equation shows that one mole of NaOH reacts with one mole of HCl.

This is written NaOH º HCl.

ºmeans "is equivalent to".

Step 4 - Work out the number of moles of the unknown substance

In step 2 we worked out the number of moles of the known substance, and in step 3 we worked out the link between the known and unknown materials. We can now put these together to work out the number of moles of the unknown substance. In the case of NaOH and HCl above, there are the same number of moles of each. So the number of moles of the unknown substance is just the same as that worked out in step 2 for the known substance.

Be careful to get the right maths here - sometimes there will be the same number of moles, sometimes there will be more (eg × 2) and sometimes there will be fewer moles (eg × ½). This is where good labeling helps out. Look at your answer to step 2. What chemical is involved? Now look at the equation for the other substance. If there are more moles of this than that in step 2, we need to scale up (eg × 2). If there are less moles of this than the other, we need to scale down (eg × ½).

Step 5 - Convert number of moles to concentration

The number of moles worked out in step 4 is the number of moles present in the volume we used - that is the volume of the pipette. We are generally asked for the concentration of the unknown material. You need to remember that concentration is measured in moles per decimetre cubed (that is moles divided by decimetre cubed). Convert the pipette volume into decimetre cubed by dividing cm³ by 1000. Then divide the number of moles worked out in step 4 by the volume we have just calculated.

Check your answer looks sensible and has the correct number of significant figures. Concentrations can vary up to around 20 M. They are usually up to 2 M, so if you get 140 M something has gone wrong!

You may be asked to give your answer in g dm^-3. This is done by multiplying the previous answer by the molar mass. This is fairly obvious if you think about a simple example. For example, NaOH has a molar mass of 40 g mol^-1. A 1 M solution contains 1 mol dm^-3 and 1 mol has a mass of 40 g, so it must be 40 g dm^-3. A 2 M solution contains 2 mol dm^-3 and a mole still has a mass of 40 g mol^-1. So this has 2 mol or 80 g dm^-3.

Example

25.0 cm³ of barium hydroxide solution was titrated with 0.0200 M hydrochloric acid. The titres were 15.00 cm³, 14.10 cm³ and 14.30 cm³. What is the concentration of the barium hydroxide in mol dm^-3 and g dm^-3?

Average titre, ignoring 15.0 cm³ = (14.10 + 14.30)/2 = 14.20 cm³

Number of moles of HCl = 14.20/1000 × 0.0200

Ba(OH)₂(aq) + 2HCl(aq) ⇒ BaCl₂(aq) + 2H₂O(l)

Ba(OH)₂ º 2HCl

There is less Ba(OH)₂ than HCl

\number of moles of Ba(OH)₂ in 25.0 cm³ = ½ × 14.20/1000 × 0.0200

\concentration of Ba(OH)₂ = (½ × 14.20/1000 × 0.0200)/0.025 = 0.0057 mol dm^-3

Ba(OH)₂ = 171

\concentration of Ba(OH)₂ = 0.0057 × 171 = 0.97 g dm^-3

You can get some further practice with exercise 1 and exercise 2.

Now see if you can apply your knowledge in slightly different situations - try problem 1 and problem 2.