The Ionic Product of Water, Kw (Oxford AQA International A Level Chemistry)
Revision Note
Written by: Alexandra Brennan
Reviewed by: Stewart Hird
Ionic Product of Water Calculations
In all aqueous solutions, some water molecules dissociate into protons and hydroxide ions
An equilibrium is established:
H2O (l) ⇌ H+ (aq) + OH- (aq)
Kw is the ionic product of water
It is the equilibrium constant for the dissociation of water at 298 K
Its value is 1.00 x 10-14 mol2 dm-6
For the ionisation of water, the equilibrium expression to find Kw is:
H2O (l) ⇌ H+ (aq) + OH- (aq)
Kw =
As the extent of ionisation is very low, only small amounts of H+ and OH- ions are formed
The concentration of H2O can therefore be regarded as constant and removed from the Kw expression
The equilibrium expression therefore becomes:
Kw = [H+] [OH-]
As the [H+] = [OH-] in pure water, the equilibrium expression can be further simplified to:
Kw = [H+]2
Worked Example
Calculate the concentration of H+ in pure water, using the ionic product of water.
Kw = 1 × 10−14 mol2 dm-6
Answer:
Step 1: Write down the equation for the partial dissociation of water:
In pure water, the following equilibrium exists:
H2O (l) ⇌ H+ (aq) + OH- (aq)
Step 2: Write down the equilibrium expression to find Kw:
Kw =
Step 3: Simplify the expression:
Since the concentration of H2O is constant, this expression can be simplified to:
Kw = [H+] [OH-]
Step 4: Further simplify the expression:
The ratio of H+ to OH- is 1:1
The concentration of H+ and OH- is, therefore, the same and the equilibrium expression can be further simplified to:
Kw = [H+]2
Step 5: Rearrange the equation to find [H+]:
[H+] =
Step 6: Substitute the values into the expression to find Kw:
[H+] =
[H+] = 1.00 x 10-7 mol dm-3
The effect of temperature on Kw
The dissociation of water to form hydrogen and hydroxide ions is an endothermic process so absorbs heat energy
H2O (l) ⇌ H+ (aq) + OH- (aq)
If temperature is increased, the forward reaction will be favoured to counteract the change and lower the temperature
Th equilibrium will shift to the right and more H+ and OH- ions will be formed causing the value of Kw to increase
So, as temperature increases the value of Kw increase
If the value of Kw increases, the pH decreases as shown below:
T (°C)
Kw (mol2 dm-6)
pH
0
0.114 x 10-14
7.47
10
0.293 x 10-14
7.27
20
0.681 x 10-14
7.08
25
1.008 x 10-14
7.00
Calculating the pH of a strong base
Strong bases are completely ionised in solution:
BOH (aq) → B+ (aq) + OH- (aq)
Therefore, the concentration of hydroxide ions [OH-] is equal to the concentration of base [BOH]
Even strong alkalis have small amounts of H+ in solution which is due to the ionisation of water
The concentration of OH- in solution can be used to calculate the pH using the ionic product of water
Once the [H+] has been determined, the pH of the strong alkali can be found using pH = -log[H+]
Similarly, the ionic product of water can be used to find the concentration of OH- ions in solution if [H+] is known, simply by dividing Kw by the [H+]
Worked Example
Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOH at 25 °C.
Kw at 25 °C = 1 × 10−14 mol2 dm-6
Sodium hydroxide is a strong base which ionises as follows:
NaOH (aq) → Na+ (aq) + OH- (aq)
Answer:
The pH of the solution is:
[H+] =
[H+] = = 6.66 x 10-14
pH = -log[H+]
pH= -log 6.66 x 10-14 = 13.17
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