The ionic product of water, Kw (A-level only) — AQA A-Level Chemistry
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The ionic product of water, Kw (A-level only) explained
Water ionises to a tiny extent: H₂O ⇌ H⁺ + OH⁻.
Read the full explanation
Writing an ordinary equilibrium constant would put [H₂O] on the bottom, but water is in such vast excess that its concentration is effectively constant, so it is absorbed into the constant to give the ionic product Kw = [H⁺][OH⁻], with units mol² dm⁻⁶. At 298 K, Kw = 1.00 × 10⁻¹⁴. In pure water the two ions form in equal numbers, so [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³ and pH = 7.00. Kw applies to every aqueous solution, which is how the pH of an alkali is found: for 0.100 mol dm⁻³ NaOH, [H⁺] = Kw/0.100 = 1.00 × 10⁻¹³ mol dm⁻³ and pH = 13.00. The dissociation is endothermic, so heating raises Kw, lowers the pH of water, and still leaves it neutral.
Your focus
- Write the expression for Kw and state its units.
- Calculate the pH of 0.100 mol dm⁻³ sodium hydroxide at 298 K, obtaining 13.00.
- Explain why water at 323 K has a pH of about 6.6 and is still neutral.
The ionic product of water, Kw (A-level only) exam tips
Marking Points
- one mark for the equation H₂O ⇌ H⁺ + OH⁻
- one mark for Kw = [H⁺][OH⁻] with units mol² dm⁻⁶
- one mark for [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³ in pure water at 298 K
- one mark for explaining that Kw rises with temperature because the dissociation is endothermic
- one mark for using [H⁺] = Kw/[OH⁻] to find the pH of a strong base
Examiner Tips
- 💡For any strong base, find [OH⁻] first, then divide Kw by it; never take the log of [OH⁻] directly.
- 💡Remember that Kw has units, and that they come from multiplying two concentrations together.
Common Mistakes
- writing Kw = [H⁺][OH⁻]/[H₂O] and still quoting the value 1.00 × 10⁻¹⁴
- giving the units of Kw as mol dm⁻³
- stating that water becomes acidic when heated, when [H⁺] = [OH⁻] still holds
- using 1.00 × 10⁻¹⁴ at a temperature other than 298 K when a different value has been supplied