- Dilution of a strong acid with distilled water increases pH.
- Dilution of a strong base with distilled water decreases pH.
The two statements above might seem a bit confusing and contradictory at first but if we work through the numbers, its true!
If you’re happy with maths, one of the quickest ways to solve a dilution problem is to think in terms of proportionality.
For a solution of a strong acid:
- If the volume increases by 10x due to dilution.
- … the [H+] concentration decreases by 10x
- … so the final pH is 1 unit higher (i.e., -log(1/10) = +1)
For a solution of a strong base:
- If the volume increases by 10x due to dilution.
- … the [OH–] concentration decreases by 10x
- … so the [H+] concentration increases by 10x (because Kw is a constant)
- … so the final pH is 1 unit lower (i.e., -log(10) = -1)
Another approach to prove these results is to work out the moles of acid or base in the sample. This approach has a few extra steps, but is equally valid.
Here is an example using a strong acid:
| Question: 900 cm3 of distilled H2O is added to 100 cm3 of 0.08 mol dm-3 HCl. This results in a 10x dilution. Calculate the change in pH that occurs during dilution. Answer: Initial pH = -log(0.08) = 1.10
H+ moles = 0.08 x (100 / 1000) = 0.008 mol Final volume = 1000 cm3 = 1 dm3 Final [H+] = 0.008 mol dm-3 Final pH = -log(0.008) = 2.10
Change in pH = +1.00 |
Here is an example using a strong base:
| Question: 900 cm3 of distilled H2O is added to 100 cm3 of 0.08 mol dm-3 NaOH at 298 K. This results in a 10x dilution. Calculate the change in pH that occurs during dilution. Kw = 1.00 x 10-14 mol2 dm-6 Answer: Initial [OH–] = 0.08 mol dm-3 Initial [H+] = Kw / [OH–] = 1.25 x 10-13 mol dm-3 Initial pH = -log(1.25 x 10-13) = 12.90
OH- moles = 0.08 x (100 / 1000) = 0.008 mol Final volume = 1000 cm3 = 1 dm3 Final [OH–] = 0.008 mol dm-3 Final [H+] = Kw / [OH–] = 1.25 x 10-12 mol dm-3 Final pH = -log(1.25 x 10-12) = 11.90
Change in pH = -1.00 |
So – we’ve shown that dilution of strong acids increases pH and dilution of strong bases decreases pH.
Interestingly – there is a limit to this as eventually the H+ or OH– derived from partial dissociation of water become more abundant than H+ or OH– derived from the acid / base. If you diluted either a strong base or a strong acid with a HUGE amount of water, you’d end up with a solution that is very close to being neutral (at 298 K, this would be pH 7.00).