Mixing pH
Use the logarithm to work out these pH values
Problem
”‹”‹”‹”‹”‹”‹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?
Getting Started
Remember that Concentration = moles / volume.
Student Solutions
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, $$[H^+] = 6.14 \times 10^{-4} mol/l$$
$$\therefore pH = -log_{10}[H^+] = 3.21\ (2\ d.p.)$$
Teachers' Resources
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
Key questions
Possible extension
Possible support