6.1.5Additional lessons
The Ion Product of Water — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 120 of 172
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 5Autoionisation
Water acts as both an acid and a base, so it dissociates slightly into ions.
Equation
- Equation Pure water ionises slightly: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), often simplified to H₂O ⇌ H⁺ + OH⁻.
Tiny extent
- Tiny extent Only a very small fraction of water molecules is ionised, so water is a very poor conductor.
Equal ions
- Equal ions In pure water [H⁺] equals [OH⁻], so water is neutral at every temperature.
Equilibrium
- Equilibrium This is a dynamic equilibrium in which ions form and recombine continually.
Section 2 of 5The Ion Product
The Ion Product
Definition
- Definition Kw = [H⁺][OH⁻], the ion product constant of water, with water itself omitted from the expression.
Value
- Value At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶, so [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³ in pure water.
Temperature
- Temperature Ionisation is endothermic, so Kw rises with temperature and the pH of pure water falls below 7 when it is hot.
Still neutral
- Still neutral Hot pure water has a lower pH than 7 but remains neutral, since [H⁺] still equals [OH⁻].
| Solution | [H⁺] versus [OH⁻] |
|---|---|
| Acidic | [H⁺] > [OH⁻] |
| Neutral | [H⁺] = [OH⁻] |
| Alkaline | [H⁺] < [OH⁻] |
Section 3 of 5Using Kw
1
In any aqueous solution at 298 K the product [H⁺][OH⁻] always equals 1.00 × 10⁻¹⁴.
2
Adding an acid raises [H⁺], so [OH⁻] falls to keep the product constant.
3
Adding an alkali raises [OH⁻], so [H⁺] falls in the same way.
4
To find [H⁺] in an alkali, divide Kw by the hydroxide ion concentration.
5
Then take −log₁₀[H⁺] to obtain the pH, and remember that this only works at 298 K.
Section 4 of 5Worked Example
Question
- A solution of NaOH has [OH⁻] = 1.0 × 10⁻³ mol dm⁻³ at 298 K.
- Calculate [H⁺] and pH.
- [H⁺] = Kw / [OH⁻] = 1.0 × 10⁻¹⁴ / 1.0 × 10⁻³ = 1.0 × 10⁻¹¹ mol dm⁻³.
- pH = −log₁₀(1.0 × 10⁻¹¹) = 11.00.
Answer [H⁺] = 1.0 × 10⁻¹¹ mol dm⁻³; pH = 11.00
Section 5 of 5Exam Checklist & Tips
6.1.5 – Key Points for the Exam
- Kw The ion product of water, [H⁺][OH⁻], equal to 1.00 × 10⁻¹⁴ at 298 K.
- Autoionisation The reaction of two water molecules to form H₃O⁺ and OH⁻.
- Neutral The condition [H⁺] = [OH⁻].
- Hydroxide ion OH⁻, the ion that makes a solution alkaline.
Common Mistakes
- Stating that pure water is always pH 7 whatever the temperature.
- Thinking that an acidic solution contains no OH⁻.
- Including [H₂O] in the Kw expression.
- Using Kw = 10⁻¹⁴ at temperatures other than 298 K without being told.
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