8.1.2Acids and Bases
The Ionic Product of Water — AP CHEMISTRY Revision Notes & Topical Questions
5 min read 5 parts Lesson 17 of 21 Unit quiz
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.
Part 1 of 5Autoionization of Water
Even pure water contains a tiny concentration of ions, formed when water molecules transfer protons to each other.
Autoionization of Water
Equilibrium
- Equilibrium 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), often simplified to H₂O(l) ⇌ H⁺(aq) + OH⁻(aq).
Expression
- Expression Liquid water is omitted from the equilibrium expression, so Kw = [H₃O⁺][OH⁻].
Value at 25 °C
- Value at 25 °C Kw = 1.0 × 10⁻¹⁴, so pure water has [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M.
Applies to all solutions
- Applies to all solutions The product [H₃O⁺][OH⁻] equals Kw in any dilute aqueous solution, not only in pure water.
Part 2 of 5Linking the Quantities
Taking negative logarithms of Kw gives simple relationships between pH and pOH.
pH
- pH pH = −log[H₃O⁺], so pH 3 means a concentration of 1 × 10⁻³ M.
pOH
- pOH pOH = −log[OH⁻], which is large when hydroxide concentration is small.
pKw
- pKw pH + pOH = pKw, which equals 14.00 at 25 °C.
Neutral solution
- Neutral solution [H₃O⁺] = [OH⁻], which gives pH = 7.00 only at 25 °C.
Acidic and basic
- Acidic and basic Acidic solutions have [H₃O⁺] greater than [OH⁻], while basic solutions have the reverse.
| Solution at 25 °C | [H₃O⁺] compared with [OH⁻] | pH |
|---|---|---|
| Acidic | Greater | Below 7 |
| Neutral | Equal | 7.00 |
| Basic | Smaller | Above 7 |
Part 3 of 5Temperature and Conjugate Pairs
Because Kw is an equilibrium constant, it responds to temperature and connects to Ka and Kb.
1
The ionization of water absorbs energy, so it is an endothermic process.
2
Raising the temperature shifts the equilibrium towards more ions, so Kw increases.
3
With more H₃O⁺ and OH⁻ present, the pH of pure water falls below 7, yet the water stays neutral because the two concentrations are still equal.
4
For any conjugate acid-base pair in water, Ka × Kb = Kw, which is 1.0 × 10⁻¹⁴ at 25 °C.
5
Therefore a stronger acid has a weaker conjugate base, and pKa + pKb = pKw.
Part 4 of 5Worked Example
Question
- A solution at 25 °C has [H₃O⁺] = 3.2 × 10⁻⁴ M.
- Calculate [OH⁻], pOH and pH.
- [OH⁻] = Kw ÷ [H₃O⁺] = 1.0 × 10⁻¹⁴ ÷ 3.2 × 10⁻⁴ = 3.1 × 10⁻¹¹ M.
- pH = −log(3.2 × 10⁻⁴) = 3.49.
- pOH = 14.00 − 3.49 = 10.51, which agrees with −log(3.1 × 10⁻¹¹).
Answer [OH⁻] = 3.1 × 10⁻¹¹ M, pH = 3.49, pOH = 10.51 (acidic solution)
Part 5 of 5Exam Checklist & Tips
8.1.2 – Key Points for the Exam
- Autoionization The reaction in which water molecules transfer protons to each other to form H₃O⁺ and OH⁻.
- Kw The ionic product of water, equal to [H₃O⁺][OH⁻].
- pOH The negative logarithm of the hydroxide ion concentration.
- Neutral solution A solution in which [H₃O⁺] equals [OH⁻].
- pKw The negative logarithm of Kw, equal to 14.00 at 25 °C.
Common Mistakes
- Stating that neutral always means pH 7, when this holds only at 25 °C.
- Including the concentration of liquid water in the Kw expression.
- Adding pH and pOH to get 14 at temperatures other than 25 °C without noting that pKw changes.
- Forgetting that a negative logarithm of a small number gives a positive pH value.
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