The Autoionization of Water and Kw

The autoionization of water is the reaction in which two water molecules transfer a proton to form a hydronium ion and a hydroxide ion, 2H₂O ⇌ H₃O⁺ + OH⁻, and its equilibrium constant is the ion product of water, K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. In brief: in pure water [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/L, so pH = pOH = 7; in any aqueous solution at 25 °C, pH + pOH = 14.00, and if [H⁺] rises, [OH⁻] falls in proportion.

Water is both a weak acid and a weak base, and its tiny self-ionization links the concentrations of H⁺ and OH⁻ in every aqueous solution. This article explains the equilibrium, shows how K_w changes with temperature, defines neutral, acidic and basic solutions, works through calculations of [OH⁻] from [H⁺], shows why neutral water is not always pH 7, and gives some history.

What is the autoionization equilibrium?

Even the purest water conducts a very small current, because a few molecules react with each other: H₂O + H₂O ⇌ H₃O⁺ + OH⁻. One molecule acts as a Brønsted-Lowry acid, the other as a base. At 25 °C only about 1 molecule in 5.5 × 10⁸ is ionized at a given moment.

What is K_w?

The equilibrium expression is K = [H₃O⁺][OH⁻] ÷ [H₂O]². Because the concentration of water is practically constant (55.5 mol/L), it is included in the constant: K_w = [H⁺][OH⁻]. At 25 °C, K_w = 1.0 × 10⁻¹⁴, and in logarithmic form pK_w = pH + pOH = 14.00. The relation holds for pure water and for all dilute aqueous solutions. For a conjugate acid-base pair, K_a × K_b = K_w.

How does K_w depend on temperature?

The autoionization is endothermic (ΔH ≈ +56 kJ/mol), so K_w increases with temperature.

Water at 100 °C is still neutral, as [H⁺] = [OH⁻], but its pH is about 6.14.

How do you classify solutions?

Worked examples

1. [OH⁻] from [H⁺]. [H⁺] = 2.5 × 10⁻³ mol/L. [OH⁻] = 1.0 × 10⁻¹⁴ ÷ 2.5 × 10⁻³ = 4.0 × 10⁻¹² mol/L. The solution is acidic.
2. [H⁺] from [OH⁻]. 0.010 mol/L NaOH: [H⁺] = 1.0 × 10⁻¹⁴ ÷ 0.010 = 1.0 × 10⁻¹² mol/L, so pH = 12.00.
3. pOH. A solution has pH 4.30, so pOH = 14.00 − 4.30 = 9.70 and [OH⁻] = 10^−9.70 = 2.0 × 10⁻¹⁰ mol/L.

4. Neutral at 37 °C. K_w = 2.4 × 10⁻¹⁴. [H⁺] = √(2.4 × 10⁻¹⁴) = 1.55 × 10⁻⁷ mol/L, pH 6.81. Blood at pH 7.4 is therefore slightly basic relative to neutral at body temperature.
5. Dilute strong acid. 1.0 × 10⁻⁸ mol/L HCl is so dilute that water’s own H⁺ matters: x² − 1.0 × 10⁻⁸x − 1.0 × 10⁻¹⁴ = 0 gives x = 1.05 × 10⁻⁷ mol/L, pH 6.98, not 8.
6. Number of ions. In 1.00 L of pure water at 25 °C there are 1.0 × 10⁻⁷ × 6.022 × 10²³ = 6.0 × 10¹⁶ H⁺ ions.

Why does K_w matter?

• It links pH and pOH, so a base can be described by pH as well.
• It explains the acidity or basicity of salts (hydrolysis) through K_aK_b = K_w.
• It sets the limits of the pH scale in water.
• Natural waters and body fluids depend on temperature-dependent equilibria.

How was K_w determined?

Friedrich Kohlrausch and Ernst Heydweiller measured the conductivity of very pure water in 1894 to 1904 and found that it is not zero, which shows that water ionizes slightly. Their measurements led to the value of the ion product near 10⁻¹⁴ at 25 °C. Later electrochemical methods, including cells without a liquid junction, gave accurate values at different temperatures.

Which mistakes are common?

Mistake 1: Using K_w = 1.0 × 10⁻¹⁴ at all temperatures

The value is valid at 25 °C only.

Mistake 2: Saying that water at 50 °C is acidic because its pH is below 7

It is neutral, since [H⁺] = [OH⁻].

Mistake 3: Forgetting that [OH⁻] is never zero in acid

Every aqueous solution contains both ions.

Mistake 4: Ignoring water’s contribution in very dilute solutions

Below about 10⁻⁶ mol/L, include the H⁺ from water.