Neutralisation calculations

How much acid does it take?

An acid and a base neutralise each other in a fixed ratio. If we know the amounts, we can find how much of one will neutralise the other. We use moles.

The concentration of a solution is measured in moles per litre, with the symbol $\text{M}$, or mol/L. A $0.10\ \text{M}$ solution has $0.10\ \text{mol}$ of solute in every litre.

One to one reactions

For $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$, one mole of acid reacts with one mole of base. At the point of neutralisation, the moles are equal.

where $M$ is the concentration and $V$ is the volume (in the same unit for both, such as mL).

Example 1. What volume of $0.10\ \text{M}$ HCl neutralises $25\ \text{mL}$ of $0.10\ \text{M}$ NaOH?

Example 2. What volume of $0.25\ \text{M}$ HCl neutralises $20\ \text{mL}$ of $0.50\ \text{M}$ NaOH?

Reactions that are not one to one

For $\text{H}_2\text{SO}_4 + 2\,\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\,\text{H}_2\text{O}$, one mole of acid needs two moles of base. So we find the moles of base first, and halve them.

Example 3. What volume of $0.050\ \text{M}$ $\text{H}_2\text{SO}_4$ neutralises $24\ \text{mL}$ of $0.10\ \text{M}$ NaOH?

• Moles of NaOH: $0.10 \times 0.024 = 0.0024\ \text{mol}$.
• Moles of $\text{H}_2\text{SO}_4$: $0.0024 \div 2 = 0.0012\ \text{mol}$.
• Volume: $\dfrac{0.0012}{0.050} = 0.024\ \text{L} = 24\ \text{mL}$.

A titration

In a titration, we add a solution of known concentration from a burette to a measured volume of another solution, with an indicator. The colour changes at the end point, when the acid is just neutralised. From the volume used, we find the unknown concentration.

The pH changes very slowly at first, and then jumps by many units for a tiny addition near the end point. This is why a single drop changes the colour of the indicator so suddenly.

Practise with new numbers

Try it in Python

This program does the three examples.

The next program draws the pH curve of a titration of $25\ \text{mL}$ of $0.08\ \text{M}$ HCl with $0.10\ \text{M}$ NaOH.

You should find that doubling the base's concentration halves the volume needed to reach the end point.

Remember these

• moles = concentration (mol/L) times volume (L).
• For a one to one reaction, $M_1 V_1 = M_2 V_2$ at neutralisation.
• If the ratio is not one to one, find the moles first and use the ratio from the equation.
• The pH jumps sharply near the end point of a titration.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How many moles of sodium hydroxide are there in $80\ \text{mL}$ of a $0.25\ \text{M}$ solution?

2. What volume of $0.40\ \text{M}$ hydrochloric acid neutralises that sodium hydroxide?

3. In a titration, what is the end point?

4. Why can we not use $M_1 V_1 = M_2 V_2$ directly for sulphuric acid and sodium hydroxide?