The Ionic Product of Water (College Board AP® Chemistry): Study Guide
The ionic product of water
The autoionization constant of water, Kw
pH, pOH, [H3O+] and [OH–] are all related to one another
Water autoionizes: a very small fraction of water molecules transfer a proton to other water molecules, forming hydronium 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–] = 1.0 × 10-14 at 25 °C
In a neutral solution [H3O+] = [OH–]
In pure water, [H3O+] can be calculated as √Kw
At 25 °C the value of Kw is 1.0 × 10-14
Taking the negative log for the equilibrium will give
pKw = pH + pOH = 14.00 at 25 °C
How does temperature affect Kw?
The ionization 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 favored
This causes an increase in the concentrations of the hydronium and hydroxide ions
This leads to the magnitude of Kw increasing
Therefore, the pH will decrease
Increasing the temperature decreases the pH of pure water, but the water stays neutral because [H3O+] = [OH–] at every temperature
Decreasing the temperature increases the pH of pure water, and the water is still neutral
Examiner Tips and Tricks
Kw = 1.0 × 10−14 and pKw = 14.00 only at 25 °C. At another temperature, you should use the Kw you are given.
In pure water:
[H3O+] = √Kw
Kw = 3.0 × 10−14 at 40 °C
This gives:
[H3O+] = 1.7 × 10−7 M
pH = 6.76
The water is still neutral, because [H3O+] = [OH−].
Kw and the pH of pure water at different temperatures
Temperature (°C) | Kw | pH of pure water |
|---|---|---|
0 | 1.1 × 10-15 | 7.48 |
10 | 2.9 × 10-15 | 7.27 |
25 | 1.0 × 10-14 | 7.00 |
40 | 3.0 × 10-14 | 6.76 |
50 | 5.5 × 10-14 | 6.63 |
100 | 5.1 × 10-13 | 6.15 |
As temperature increases, Kw increases, so the pH of pure water decreases
The water stays neutral because [H3O+] = [OH–]
Relationship between H+ (H3O+), OH–, pH and pOH
![Cycle diagram linking pH, H+(aq), OH−(aq) and pOH: [H+] = 10 to the power −pH; pH = −log10[H+]; [OH−] = Kw ÷ [H+]; [H+] = Kw ÷ [OH−]; pOH = −log10[OH−]; [OH−] = 10 to the power −pOH; pH + pOH = pKw](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2023/07/diagram-to-show-the-relationship-between-h-and--ph-oh--and-poh.webp)
To make a conversion, follow the arrow and equation given, so to convert OH– (aq) to pOH use pOH = −log [OH–]
Worked Example
Convert between [H3O+], [OH–], pH and pOH.
Complete the following table for solutions at 25 °C, giving pH and pOH values to 2 decimal places.
[H3O+] | [OH–] | pH | pOH |
|---|---|---|---|
2.40 × 10-4 |
|
|
|
| 4.60 × 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 × 10-4 |
= 4.17 × 10-11 | −log (2.40 × 10-4) = 3.62 | 14.00 − pH = 10.38 |
= 2.17 × 10-10 | 4.60 × 10-5 | 14.00 − pOH = 9.66 | −log (4.60 × 10-5) = 4.34 |
10-10.36 = 4.37 × 10-11 | = 2.29 × 10-4 | 10.36 | 14.00 − pH = 3.64 |
= 1.62 × 10-11 | 10-3.21 = 6.17 × 10-4 | 14.00 − pOH = 10.79 | 3.21 |
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