The Ionic Product of Water
The autoionization constant of water, Kw
- The pH, pOH [H3O+] and [OH-] are all related to one another
- An equilibrium exists in water, where a few water molecules dissociate into proton and hydroxide ions
2H2O (l) ⇌ H3O+ (aq) + OH– (aq)
OR
H2O (l) ⇌ H+ (aq) + OH– (aq)
- The equilibrium expression for this is:
Kw = [H3O+] [OH–] = 1.00 x 10-14
- In a neutral solution [H3O+] = [OH–]
- [H3O+] can be calculated a
- At 25 °C the value of Kw is 1.00 x 10-14
- Since the concentration of the H3O+ and OH- ions is very small, the concentration of water is considered to be a constant
- This means that the expression can be rewritten as:
Kw = [H3O+] [OH-]
-
- Where Kw = 1.00 10-14 at 25 °C
- Where Kw = 1.00 10-14 at 25 °C
- Taking the negative log for the equilibrium will give
pKw = pH + pOH = 14
How does temperature affect Kw?
- The ionisation of water is an endothermic process
2H2O (l) ⇌ H3O+ (aq) + OH- (aq)
- In accordance with Le Châtelier's principle, an increase in temperature will result in the forward reaction being favoured
- This causes an increase in the concentration of the hydrogen and hydroxide ions
- This leads to the magnitude of Kw increasing
- Therefore, the pH will decrease
- Increasing the temperature, decreases the pH of water (becomes more acidic)
- Decreasing the temperature, increases the pH of water (becomes more basic)
Graph to show how Kw changes with temperature
As temperature increases, Kw increases so pH decreases
Relationship between H+ (H3O+) , OH–, pH and pOH
To make a conversion, follow the arrow and equation given, so to convert OH– (aq) to pOH use pOH = -log10[OH–]
Worked example
It is good practise to convert [between H+] (or [H3O+]), [OH–], pH and pOH.
Complete the following table giving your answers to 2 decimal places.
[H3O+] | [OH–] | pH | pOH |
2.40 x 10-4 | |||
4.60 x 10-5 | |||
10.36 | |||
3.21 |
Answer - As shown below there are different ways to arrive at the correct answer
[H3O+] | [OH–] | pH | pOH |
2.40 x 10-4 | = 4.17 x 10-11 | -log (2.40 x 10-4) = 3.62 | 14 - pH = 10.38 |
= 2.17 x 10-10 | 4.60 x 10-5 | 14 - pOH = 9.66 | -log (4.60) = 4.34 |
10-10.36 = 4.37 x 10-11 | = 2.29 x 10-4 | 10.36 | 14 - pH = 3.64 |
= 1.62 x 10-11 | 10-3.21 = 6.17 x 10-4 | 14 - pOH = 10.79 | 3.21 |