Ionic Product of Water (Kw): Why Neutral pH Is Not Always 7

By Aniket Bhardwaj · 24 October 2026 · Updated 24 October 2026 · Class 11 Chemistry

Most students learn one fact about water: “pH of pure water is 7.” Then a board paper asks for the pH of pure water at 100 °C and the answer is not 7. This is a favourite trap because the question looks too easy. The idea behind it is the ionic product of water, K_w, and the fact that its value depends on temperature.

A comparison shows that neutral pH shifts from 7.00 at 25 °C to 6.13 at 100 °C as Kw rises from 1.0 × 10⁻¹⁴ to 5.5 × 10⁻¹³. — Neutral pH changes with temperature as water’s ionic product rises.

What K_w is

Water ionises very slightly: H₂O ⇌ H⁺ + OH⁻. (In full, H⁺ is H₃O⁺, but the short form is fine for school problems.) The product of the two ion concentrations in any aqueous solution is a constant at a given temperature.

K_w = [H⁺] × [OH⁻]
pK_w = −log K_w = pH + pOH
Neutral water: [H⁺] = [OH⁻] = √K_w, so neutral pH = ½ pK_w

[H⁺] and [OH⁻] are in mol L⁻¹. K_w has no unit in most school books. At 25 °C, K_w = 1.0 × 10⁻¹⁴, so pK_w = 14 and neutral pH = 7. That is where the well-known “pH + pOH = 14” comes from. It is a result for 25 °C, not a law for all temperatures.

Why does K_w change? The ionisation of water absorbs heat (it is endothermic). By Le Chatelier’s principle, raising the temperature pushes the equilibrium to the right, so more H⁺ and OH⁻ form and K_w increases. As K_w rises, pK_w falls.

Temperature	K_w used in the examples	pK_w	Neutral pH
25 °C	1.0 × 10⁻¹⁴	14.00	7.00
100 °C	5.5 × 10⁻¹³	12.26	6.13

The 100 °C value is a rounded textbook figure, and books differ slightly (some use 1.0 × 10⁻¹² for easy arithmetic). Always use the K_w your question gives. If it gives none, assume 25 °C.

Worked example 1: pH of pure water at 100 °C

Question. K_w = 5.5 × 10⁻¹³ at 100 °C. Find the pH of pure water and say whether it is acidic.

In pure water [H⁺] = [OH⁻] = √(5.5 × 10⁻¹³) = 7.42 × 10⁻⁷ M.

pH = −log(7.42 × 10⁻⁷) = 6.13.

Answer. pH = 6.13. The water is still neutral, because [H⁺] = [OH⁻]. Neutral does not mean pH = 7; it means equal concentrations of H⁺ and OH⁻.

Worked example 2: is pH 6.5 acidic or basic at 100 °C?

Use K_w = 5.5 × 10⁻¹³, so pK_w = 12.26 and neutral pH = 6.13.

A solution with pH 6.5 has [H⁺] = 10^−6.5 = 3.16 × 10⁻⁷ M.

pOH = 12.26 − 6.50 = 5.76, so [OH⁻] = 10^−5.76 = 1.74 × 10⁻⁶ M.

Here [OH⁻] is larger than [H⁺], so the solution is basic at 100 °C, although most students would call pH 6.5 “acidic” because it is below 7. The rule is: compare the pH with neutral pH at that temperature, not with 7.

Worked example 3: 0.010 M NaOH at 100 °C

NaOH is a strong base, so [OH⁻] = 0.010 M and pOH = 2.00.

Wrong way: pH = 14 − 2 = 12.

Right way at 100 °C: [H⁺] = K_w ÷ [OH⁻] = 5.5 × 10⁻¹³ ÷ 0.010 = 5.5 × 10⁻¹¹ M. pH = −log(5.5 × 10⁻¹¹) = 10.26.

Check: pH = pK_w − pOH = 12.26 − 2.00 = 10.26. Both routes agree.

Worked example 4: finding K_w from a pH

Question. At a certain temperature pure water has pH 6.5. Find K_w and pK_w.

Neutral: [H⁺] = [OH⁻] = 10^−6.5 = 3.16 × 10⁻⁷ M.

K_w = (3.16 × 10⁻⁷)² = 1.0 × 10⁻¹³. So pK_w = 13.00 (also 2 × 6.5 = 13).

Since K_w is larger than 1.0 × 10⁻¹⁴, this temperature is above 25 °C, which matches the idea that K_w grows on heating.

Worked example 5: a strong acid at 100 °C

Question. Find pH and [OH⁻] of 0.010 M HCl at 100 °C (K_w = 5.5 × 10⁻¹³).

HCl is a strong acid, so [H⁺] = 0.010 M and pH = 2.00. The acid concentration is so much larger than the 7.4 × 10⁻⁷ M from water that water’s own ions can be ignored.

[OH⁻] = K_w ÷ [H⁺] = 5.5 × 10⁻¹³ ÷ 0.010 = 5.5 × 10⁻¹¹ M, so pOH = 10.26.

Check: pH + pOH = 2.00 + 10.26 = 12.26 = pK_w. For a strong acid, the pH itself does not depend on temperature in this simple model; only the pOH and the neutral point change.

A simple way to think about it

Imagine a seesaw. In neutral water the two ions are balanced: [H⁺] = [OH⁻]. Adding acid raises [H⁺] and pushes [OH⁻] down so that the product stays at K_w. Heating water makes the whole seesaw heavier on both sides, so the product K_w becomes larger, but the balance point is still neutral. The pH scale from 0 to 14 is therefore a 25 °C picture. At 100 °C the scale effectively runs from 0 to about 12.26, with neutral at about 6.13.

This is also why a pH meter used for hot solutions is calibrated or temperature-compensated: the same [H⁺] gives the same pH, but the meaning of “neutral” moves with temperature.

Common mistakes

Writing pH + pOH = 14 at every temperature. The sum is pK_w, which is 14 only at 25 °C.
Calling pure hot water acidic because its pH is below 7. It is neutral if [H⁺] = [OH⁻].
Thinking a pH below 7 always means an acid. Compare with ½ pK_w for that temperature.
Taking K_w as 10⁻⁷. That is [H⁺] of pure water at 25 °C, not K_w.
Forgetting the square root when finding [H⁺] of pure water from K_w.

Exam relevance

Question type	What to do
pH of pure water at a given temperature	[H⁺] = √K_w, then −log
Is the solution acidic, basic or neutral?	Compare [H⁺] with [OH⁻], or pH with ½ pK_w
pH of a strong base at a non-standard temperature	pH = pK_w − pOH
Assertion–reason on pH of water on heating	K_w rises, pH falls, water stays neutral
No temperature given	Assume 25 °C and K_w = 1.0 × 10⁻¹⁴

Read the question carefully: if it states a temperature, it expects you to use the K_w given for that temperature.

After working out pOH or [H⁺] by hand, enter the value in the pH tool to check the last step.

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