The Ionic Product of Water, Kw (Oxford AQA International A Level (IAL) Chemistry): Revision Note
Exam code: 9622
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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