The Ionic Product of Water (AQA A Level Chemistry): Revision Note
Exam code: 7405
The Ionic Product of Water
In all aqueous solutions, an equilibrium exists in water where a few water molecules dissociate into protons and hydroxide ions
We can derive an equilibrium constant for the reaction:
This is a specific equilibrium constant called the ionic product for water
The product of the two ion concentrations is always 1 x 10-14 mol2 dm-6
This makes it straightforward to see the relationship between the two concentrations and the nature of the solution:
[H+] & [OH–] Table
![[H+] and [OH-] table, downloadable IB Chemistry revision notes](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2021/06/8.1.8-H-and-OH-table.png)
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 |
pH of Strong Bases
Strong bases
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 founding 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
pH calculations of a strong alkaliQuestion 1: Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOHQuestion 2: Calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50
Answer
Sodium hydroxide is a strong base which ionises as follows:
NaOH (aq) → Na+ (aq) + OH- (aq)
Answer 1:
The pH of the solution is:
[H+] = Kw ÷ [OH-]
[H+] = (1 x 10-14) ÷ 0.15 = 6.66 x 10-14
pH = -log[H+]
= -log 6.66 x 10-14 = 13.17
Answer 2
Step 1: Calculate hydrogen concentration by rearranging the equation for pH
pH = -log[H+]
[H+]= 10-pH
[H+]= 10-10.50
[H+]= 3.16 x 10-11 mol dm-3
Step 2: Rearrange the ionic product of water to find the concentration of hydroxide ions
Kw = [H+] [OH-]
[OH-]= Kw ÷ [H+]
Step 3: Substitute the values into the expression to find the concentration of hydroxide ions
Since Kw is 1 x 10-14 mol2 dm-6,
[OH-]= (1 x 10-14) ÷ (3.16 x 10-11)
[OH-]= 3.16 x 10-4 mol dm-3
Worked Example
What is the pH of a solution of hydroxide ions of concentration 1.0 × 10−3 mol dm−3 ?Kw = 1 × 10−14 mol2 dm-6
A. 3.00
B. 4.00
C. 10.00
D. 11.00
Answer
The correct option is D.
Since Kw = [H+] [OH–], rearranging gives [H+] = Kw ÷ [OH–]
The concentration of [H+] is (1 × 10−14) ÷ (1.0 × 10−3) = 1.0 × 10−11 mol dm−3
[H+]= 10-pH
So the pH = 11.00
Examiner Tips and Tricks
Always give the pH to two decimal places.
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