Water is amphiprotic, which means it can hand a proton to itself. It happens in about two molecules in every billion — and that tiny reaction is what fixes the pH scale, explains alkalis, and quietly moves the neutral point when you heat the beaker.
📚 What you need to know
Water self-ionises: H2O(l) ⇌ H+(aq) + OH–(aq).
Kw = [H+][OH–] = 1.00 × 10–14 at 298 K.
Water is left out of the expression because its concentration is effectively constant.
In any aqueous solution the two concentrations multiply to Kw — push one up and the other falls.
Self-ionisation is endothermic, so heating raises Kw and lowers the pH of pure water — which stays neutral.
Water reacting with itself
Every so often two water molecules collide hard enough for one to hand a proton to the other. One becomes H3O+, the other OH–. It is proton transfer, with water playing both parts.
Self-ionisation of water
2H2O(l) ⇌ H3O+(aq) + OH–(aq) often written more simply as H2O(l) ⇌ H+(aq) + OH–(aq)
Write the equilibrium expression for it and something familiar happens. Water is the solvent, present in vast excess, so its concentration barely changes and is treated as a constant. Fold that constant into K and you get a new constant of your own.
Where Kw comes from
Kc = [H+][OH–] / [H2O] so Kc × [H2O] = Kw = [H+][OH–]
This is exactly the reasoning you used for leaving solids out of an equilibrium expression. A pure liquid or a solvent in huge excess has a concentration that cannot meaningfully change, so it is absorbed into the constant rather than written as a variable.
The product is fixed; the split is not
Here is the powerful part. Kw applies to every aqueous solution, not just pure water. Add acid and [H+] shoots up — but the product must stay at 1.00 × 10–14, so [OH–] is forced down to compensate. There are always some hydroxide ions in an acid, and always some hydrogen ions in an alkali.
A schematic, not a scale drawing: the real numbers span eight powers of ten across these three boxes. The point is that the area — the product — never changes.
[H+] / mol dm–3
[OH–] / mol dm–3
pH at 298 K
Solution is
1 × 10–1
1 × 10–13
1
strongly acidic
1 × 10–3
1 × 10–11
3
acidic
1 × 10–7
1 × 10–7
7
neutral
1 × 10–11
1 × 10–3
11
alkaline
1 × 10–13
1 × 10–1
13
strongly alkaline
Getting the pH of an alkali
An alkali gives you [OH–], but the pH formula wants [H+]. Kw is the bridge between them.
The bridge
[H+] = Kw ÷ [OH–]
🧩 pH of a solution of a strong base
Work out [OH–] from the concentration of the base, remembering the formula: Ba(OH)2 gives two hydroxide ions per formula unit.
Divide Kw by [OH–] to get [H+].
Take –log10 of that.
Sanity check: the answer should be above 7.
WORKED EXAMPLE
Calculate the pH of a 0.100 mol dm–3 solution of sodium hydroxide at 298 K. (Kw = 1.00 × 10–14)
Step 1 — hydroxide concentrationNaOH is a strong base and gives one OH⁻ per formula unit.[OH⁻] = 0.100 mol dm⁻³Step 2 — cross to [H⁺][H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.100 = 1.00 × 10⁻¹³Step 3 — take the logpH = −log₁₀(1.00 × 10⁻¹³)pH = 13.00Well above 7, as an alkali must be. Note the shortcut: pOH = 1.00, so pH = 14.00 − 1.00.
pOH, the shortcut
Because the product of the two concentrations is fixed, the sum of their logarithms is fixed too. That gives an equation worth memorising.
At 298 K only
pOH = –log10[OH–] and pH + pOH = 14.00
The “14” is not a magic number. It is –log10(1.00 × 10–14), the negative logarithm of Kw. Change the temperature and Kw changes, so the 14 changes with it.
WORKED EXAMPLE
A solution at 298 K has a pH of 4.20. Calculate the concentration of hydroxide ions in it.
Step 1 — from pH to [H⁺][H⁺] = 10⁻⁴·²⁰ = 6.31 × 10⁻⁵ mol dm⁻³Step 2 — rearrange Kw[OH⁻] = 1.00 × 10⁻¹⁴ ÷ (6.31 × 10⁻⁵)[OH⁻] = 1.58 × 10⁻¹⁰ mol dm⁻³Or via pOH: 14.00 − 4.20 = 9.80, and 10⁻⁹·⁸⁰ gives the same answer. Use whichever route you trust more.
What temperature does to Kw
Breaking a bond to pull water apart costs energy, so self-ionisation is endothermic. Apply Le Chatelier: raise the temperature and the equilibrium shifts right to absorb the extra energy. Both [H+] and [OH–] increase, so their product — Kw — increases too.
At 100 °C the ionic product is about fifty times its value at room temperature, and the pH of pure water has dropped to roughly 6.1.
Read that graph carefully, because it hides the single most-missed idea in this topic. Hot pure water has a pH below 7 and is still perfectly neutral. Neutral means [H+] = [OH–], and in pure water that is true whatever the temperature. “pH 7 is neutral” is a convenient statement about 298 K, not a definition.
WORKED EXAMPLE
At 323 K, Kw = 5.48 × 10–14. Calculate the pH of pure water at this temperature and state, with a reason, whether it is acidic, neutral or alkaline.
Step 1 — use the fact that it is pure waterIn pure water every H⁺ comes with an OH⁻, so the two concentrations are equal.[H⁺]² = 5.48 × 10⁻¹⁴Step 2 — square root[H⁺] = 2.34 × 10⁻⁷ mol dm⁻³Step 3 — take the logpH = −log₁₀(2.34 × 10⁻⁷)pH = 6.63Acidic or not?still neutral[H⁺] equals [OH⁻], which is the definition of neutral. The pH is below 7 only because Kw is larger at this temperature.
WORKED EXAMPLE
State and explain the effect of cooling pure water from 298 K to 283 K on (a) the value of Kw, (b) the pH, (c) whether the water is still neutral.
(a) the constantSelf-ionisation is endothermic, so cooling shifts the equilibrium left, in the exothermic direction.Kw decreases(b) the pHFewer ions form, so [H⁺] falls.the pH rises above 7(c) neutralityyes, still neutralThe two ions are still produced in a 1 : 1 ratio, so they remain equal. Only their common value has changed.
💡 Exam tip
Quote Kwwith its temperature. The value 1.00 × 10–14 belongs to 298 K and nowhere else.
For an alkali, always go [OH–] → [H+] → pH, or use pOH. Never put [OH–] into the pH formula.
Check the formula of the base: Ca(OH)2 and Ba(OH)2 release two OH– per formula unit.
For pure water at any temperature, [H+] = √Kw.
Define neutral as [H+] = [OH–], not as “pH 7”. Examiners test this deliberately.
Explain temperature effects using Le Chatelier and the endothermic ionisation, not by assertion.
⚠️ Common mix-up
Calling hot pure water acidic because its pH is below 7. It is neutral.
Using pH + pOH = 14 at a different temperature. The sum is 14.00 only at 298 K.
Taking –log of [OH–] and calling the answer the pH.
Forgetting the 2 in Ca(OH)2, which halves the hydroxide concentration and shifts the pH by 0.3.
Thinking an acidic solution contains no OH–. It always contains some — the product is fixed.
Assuming Kw changes when you add acid. Only temperature moves it.
Up next: Strong and Weak Acids and Bases — every calculation here assumed you already knew [H+]. For a weak acid you do not, because only a fraction of it ever dissociates, and that changes everything.
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