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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

📚 IB Chemistry SL Hub / Topical Revision
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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.
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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⁻³Type
1 × 10⁻³1 × 10⁻¹¹Acidic
1 × 10⁻⁷1 × 10⁻⁷Neutral
1 × 10⁻¹¹1 × 10⁻³Alkaline

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.

  1. NaOH is a strong base that fully dissociates, so [OH⁻] = 0.020 mol dm⁻³.
  2. [H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.020 = 5.0 × 10⁻¹³ mol dm⁻³.
  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.

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