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Ask a class what the pH of pure water is and almost everyone says 7. Then ask whether boiling water is acidic. The honest answer is no, even though its pH is about 6.1. Both statements are true because "neutral" is defined by equal concentrations of H3O⁺ and OH⁻, not by the number 7. This post explains why, gives a table you can check, and works three examples where the 25 °C habit gives the wrong answer.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list the "ionization of water" and the "pH scale" under ionic equilibrium. The NCERT Class 11 Equilibrium unit (Reprint 2026-27) defines the ionic product Kw = [H⁺][OH⁻], gives 1 × 10⁻¹⁴ at 298 K, and adds a line many students skip: "The value of Kw is temperature dependent as it is an equilibrium constant." It defines a neutral solution as one where [H3O⁺] = [OH⁻]. Check the syllabus for your exam year.
Why does Kw rise with temperature?
OpenStax Chemistry 2e explains that the self-ionisation of water, 2H2O ⇌ H3O⁺ + OH⁻, "is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature". By Le Chatelier's principle, heating an endothermic equilibrium shifts it forward, so Kw increases. OpenStax gives Kw of about 5.6 × 10⁻¹³ at 100 °C, roughly 50 times its value at 25 °C.
In pure water [H3O⁺] = [OH⁻] = √Kw, so the neutral pH is always half of pKw:
neutral pH = ½ pKw
How much does neutral pH change?
The International Association for the Properties of Water and Steam publishes the reference values. Its revised release on the ionization constant of H2O (R11-24), built on the Bandura and Lvov formulation, tabulates pKw for liquid water:
0 °C: pKw = 14.95, neutral pH about 7.5
25 °C: pKw = 13.99, neutral pH 7.0
50 °C: pKw = 13.26, neutral pH 6.63
75 °C: pKw = 12.70, neutral pH 6.35
100 °C: pKw = 12.25, neutral pH about 6.1
Worked example 1: is a pH 7.00 solution neutral at 50 °C?
At 50 °C, pKw = 13.26.
[H3O⁺] = 10⁻⁷·⁰⁰ = 1.0 × 10⁻⁷ M.
[OH⁻] = Kw/[H3O⁺] = 10⁻¹³·²⁶/10⁻⁷·⁰⁰ = 10⁻⁶·²⁶ = 5.5 × 10⁻⁷ M.
[OH⁻] is about five times [H3O⁺], so the solution is slightly basic. Quick check: neutral pH at 50 °C is 6.63, and 7.00 is above it.
Worked example 2: 0.010 M NaOH at 100 °C
NaOH is a strong base, so [OH⁻] = 0.010 M and pOH = 2.00. Ignoring activity effects:
At 25 °C: pH = 13.99 − 2.00 = 11.99, which is the familiar 12.
At 100 °C: pH = 12.25 − 2.00 = 10.25.
Same solution, same [OH⁻], but the pH is almost two units lower because pKw is lower. Using pH + pOH = 14 here would give 12.00, which is wrong.
Worked example 3: hot pure water from Kw
OpenStax gives Kw = 2.4 × 10⁻¹³ at 80 °C. Then [H3O⁺] = √(2.4 × 10⁻¹³) = 4.9 × 10⁻⁷ M, and pH = −log(4.9 × 10⁻⁷) = 6.31. Below 7, and still neutral, because [OH⁻] is also 4.9 × 10⁻⁷ M.
Common mistakes and how to avoid them
"pH below 7 means acidic." Only at 25 °C. Compare with ½ pKw at the given temperature.
Using pH + pOH = 14 everywhere. NCERT writes pKw = pH + pOH = 14 for 298 K. At other temperatures use that temperature's pKw.
Thinking hot water has extra H3O⁺ only. Heating raises both ions equally.
Mixing up K and pK. Kw goes up with temperature, so pKw goes down.
Our guide to pH, buffers and ionic equilibrium covers weak acids and salts at 298 K. If ionic equilibrium keeps tripping you up and you are unsure where to begin, Fermi's AI tutor can work out what you already know and build a plan for the session.
A practice routine
Copy the five-row table and add a column for [H3O⁺] in pure water at each temperature.
NCERT Exercise 6.65 gives Kw = 2.7 × 10⁻¹⁴ at 310 K. Find the neutral pH (you should get 6.78).
Redo Worked example 2 for 0.0010 M NaOH at 50 °C.
Explain to a friend, in three sentences, why boiling water is not acidic.
Next time a question mentions a temperature, write "neutral pH = ½ pKw" in the margin before you start.
