6.1.5Additional lessons
The Ion Product of Water — IB Chemistry SL Revision Notes & Topical Questions
5 min read 6 parts Lesson 81 of 98
Part 1 of 6Self-Ionisation of Water
Kw applies to every aqueous solution, not just pure water, at a given temperature.
Equilibrium
- Equilibrium Water ionises slightly: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq), so pure water contains equal amounts of H⁺ and OH⁻.
Ion product
- Ion product Kw = [H⁺][OH⁻], and the concentration of water is so large that it is treated as constant.
Value at 298 K
- Value at 298 K Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶, so in pure water [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³.
Neutral
- Neutral A solution is neutral when [H⁺] = [OH⁻], which gives pH 7.00 only at 298 K.
Part 2 of 6Acidic, Neutral and Alkaline Solutions
Acidic, Neutral and Alkaline Solutions
Acidic
- Acidic [H⁺] is greater than [OH⁻], so [H⁺] > 1 × 10⁻⁷ mol dm⁻³ at 298 K.
Neutral
- Neutral The concentrations of hydrogen ions and hydroxide ions are exactly equal, so [H⁺] = [OH⁻].
Alkaline
- Alkaline [OH⁻] is greater than [H⁺], so [H⁺] < 1 × 10⁻⁷ mol dm⁻³ at 298 K.
Link to pOH
- Link to pOH pOH = −log[OH⁻], and at 298 K pH + pOH = 14.00.
| [H⁺] / mol dm⁻³ | [OH⁻] / mol dm⁻³ | Type |
|---|---|---|
| 1 × 10⁻³ | 1 × 10⁻¹¹ | Acidic |
| 1 × 10⁻⁷ | 1 × 10⁻⁷ | Neutral |
| 1 × 10⁻¹¹ | 1 × 10⁻³ | Alkaline |
Part 3 of 6Temperature and Kw
1
The ionisation of water is endothermic because bonds in water must be broken.
2
Raising the temperature shifts the equilibrium to the right according to Le Chatelier's principle.
3
Both [H⁺] and [OH⁻] increase, so Kw increases with temperature.
4
The pH of pure water therefore falls below 7 when heated, yet the water stays neutral because [H⁺] still equals [OH⁻].
Part 4 of 6Linking Kw, pH and pOH
Linking Kw, pH and pOH
Equation
- Equation pH + pOH = pKw, which equals 14.00 at 298 K.
Alkaline solution
- Alkaline solution For [OH⁻] = 1 × 10⁻³ mol dm⁻³, pOH = 3 and pH = 11.
Dilution of alkali
- Dilution of alkali Diluting an alkali ten times lowers its pH by one unit.
Exam use
- Exam use Use Kw to find [H⁺] first, then take the negative logarithm.
Part 5 of 6Worked Example
Question Calculate the pH of 0.020 mol dm⁻³ NaOH at 298 K, with Kw = 1.00 × 10⁻¹⁴.
- NaOH is a strong base that fully dissociates, so [OH⁻] = 0.020 mol dm⁻³.
- [H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.020 = 5.0 × 10⁻¹³ mol dm⁻³.
- pH = −log(5.0 × 10⁻¹³) = 12.30.
Answer pH = 12.30.
Part 6 of 6Exam Checklist & Tips
6.1.5 – Key Points for the Exam
- Kw The ion product constant of water, equal to [H⁺][OH⁻].
- Self-ionisation The slight dissociation of water into H⁺ and OH⁻ ions.
- pOH The negative logarithm of the hydroxide ion concentration.
- Neutral solution A solution in which [H⁺] equals [OH⁻].
Common Mistakes
- Stating that neutral always means pH 7; this is true only at 298 K.
- Using Kw = 10⁻¹⁴ at temperatures other than 298 K without being told it.
- Forgetting to take the logarithm when converting [H⁺] calculated from Kw into pH.
- Saying that hot water is acidic because its pH is below 7.
Last updated:
\n