4.5.5Unit 4. Rates, Equilibria and Further Organic Chemistry
Ionic Product of Water — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 5 parts Lesson 37 of 56 Unit quiz
Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
Section 1 of 5Kw Expression
Water ionises very slightlyH₂O(l) ⇌ H⁺(aq) + OH⁻(aq).
Definition
- Definition Kw = [H⁺][OH⁻], the ionic product of water, with units of mol² dm⁻⁶.
Value at 298 K
- Value at 298 K At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ for all aqueous solutions, whether acidic, neutral or alkaline.
Pure water
- Pure water In pure water [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³, giving a pH of 7.00 at 298 K.
Why water is omitted
- Why water is omitted The concentration of liquid water is effectively constant, so it is absorbed into the value of Kw.
Section 2 of 5pH of Strong Bases
The pH of an alkali is found by first working out [H⁺] from Kw.
1
A strong monoacidic base such as NaOH is fully dissociated, so [OH⁻] equals the concentration of the base.
2
Rearrange the ionic product to give [H⁺] = Kw / [OH⁻] at the given temperature.
3
Take −log₁₀ of [H⁺] to obtain the pH, quoted to two decimal places.
4
For Ba(OH)₂, which gives two hydroxide ions per formula unit, double the base concentration to find [OH⁻].
Section 3 of 5Effect of Temperature
Effect of Temperature
Endothermic ionisation
- Endothermic ionisation Breaking bonds in water is endothermic, so raising the temperature moves the position to the right.
Kw increases
- Kw increases At higher temperatures Kw is greater, for example about 5.5 × 10⁻¹⁴ mol² dm⁻⁶ at 50 °C.
Neutral pH falls
- Neutral pH falls Neutral water at 50 °C has [H⁺] = [OH⁻] = 2.3 × 10⁻⁷ mol dm⁻³, so pH is 6.63, not 7.
Still neutral
- Still neutral The water remains neutral because [H⁺] still equals [OH⁻], even though the pH is below 7.
Section 4 of 5Worked Example
Question Calculate the pH of 0.0100 mol dm⁻³ NaOH at 298 K, where Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.
- NaOH is a strong base, so [OH⁻] = 0.0100 mol dm⁻³.
- [H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.0100 = 1.00 × 10⁻¹² mol dm⁻³.
- pH = −log₁₀(1.00 × 10⁻¹²) = 12.00.
Answer pH = 12.00.
Section 5 of 5Exam Checklist & Tips
4.5.5 – Key Points for the Exam
- Kw Ionic product of water, [H⁺][OH⁻].
- Neutral A solution in which [H⁺] equals [OH⁻].
- Alkali A soluble base that releases OH⁻ ions in solution.
- pOH −log₁₀[OH⁻].
Common Mistakes
- Assuming pH 7 is neutral at every temperature.
- Forgetting to double [OH⁻] for Ba(OH)₂ or Ca(OH)₂.
- Using the base concentration directly as [H⁺].
- Giving Kw without units.
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