# Mixing pH

Use the logarithm to work out these pH values

”‹”‹”‹”‹”‹”‹The pH of a solution is defined using logarithms as

$$

pH = -\log_{10}[H^+],

$$

where $[H^+]$ is the concentration of $H^+$ ions in mol/l of the solution.

- Given that the pH of a beaker of pure water is 7, work out how many $H^+$ ions there are in 1 litre of the water.
- A strong acid has a pH of 2. If one litre of this acid is diluted with 1 litre of water, what is the pH of the resulting solution?
- A strong acid has a pH of 1.3. If I have 100ml of this acid, how much water needs to be added to create a solution of pH 2?
- 400ml of an acid of pH 3 is added to 300ml of an acid of pH 4. What is the resulting pH?

Make up some of your own mixing pH questions. For example:

- If I start with $1$ litre of acid of pH $1$, what happens to the pH each time I add $100$ml of water? What sort of curve results?
- Is it the case that when mixing two solutions, the resulting pH is always between the pH of the two initial solutions?

Remember that Concentration = moles / volume.

1) The initial equation can be rearranged to give: $$[H^+] = 10^{-pH}$$

The pH of water is 7, and so: $$[H^+] = 10^{-7}mol/l.$$

Thus in 1l of water there are 10$^{-7}$ moles of hydrogen ions.

2) If the pH is 2, then $[H^+] = 10^{-2} mol/l$

If this is diluted to a volume of 2 litres from the initial 1 litre, then $[H^+] = 0.5 \times 10^{-2}mol/l$.

$$\therefore pH = -log_{10}(0.5 \times 10^{-2}) = 2.30\ (2\ d.p)$$

3) If pH = 1.3, then $[H^+] = 10^{-1.3} mol/l$

Therefore, in 100ml (= 0.1l), there are $0.1 \times 10 ^{-1.3}$ moles.

We want to dilute this such that the pH = 2, which is the same as $[H^+] = 10^{-2} mol/l$.

$$\therefore \frac{0.1 \times 10^{-1.3}}{Vol} = 10^{-2}$$

$$Vol = 0.501l$$

Therefore, since we have already 100ml of solution, this requires the addition of 401ml.

4) For the first acid, pH = 3, therefore $[H^+] = 10^{-3} mol/l$.

Since there is 400ml of this acid, then the number of $H^+$ is $0.4 \times 10^{-3}$ moles

For the second acid, pH = 4, therefore $[H^+] = 10^{-4} mol/l$.

Since there is 300ml of this acid, the number of $H^+$ is $0.3 \times 10^{-4}$ moles

If these are mixed, there are a total of $4.3 \times 10^{-4}$ moles of $H^+$ in a volume of 0.7l. Therefore, $$[H^+] = 6.14 \times 10^{-4} mol/l$$

$$\therefore pH = -log_{10}[H^+] = 3.21\ (2\ d.p.)$$

### Why do this problem?

This problem provides a good exercise in linking two areas of mathematics: concentration and logarithms. Students will need to flip between the two descriptions of acidity which will require careful, clear thinking. The context is fascinating and this will encourage a 'concrete' and practical engagement with logarithms.### Possible approach

This question uses the advanced concept of logarithms but will
also use basic ideas about concentrations and mixtures. Students
will need to remember basic ideas about the meaning of a
'concentration' in order to work out the concentration of
mixtures.

### Key questions

What equations do you need to flip between the pH and the
concentration of $H^+$ ions?

How can you work out the concentration of a mixture of two
solutions?

### Possible extension

Try the next question in the sequence:
Extreme dissociation

### Possible support

You might try and get started using the pH formula with the
problem Temperature
pH