Water Autoionization Calculators

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Water autoionization (self-ionization) is the equilibrium process in which two water molecules react to form a hydronium ion (H₃O⁺) and a hydroxide ion (OH⁻): 2H₂O ⇌ H₃O⁺ + OH⁻. The equilibrium constant for this reaction at 25°C is the water dissociation constant Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴. This relationship defines the pH scale: pH = −log₁₀[H⁺]. At 25°C, neutral water has pH = 7.0 (equal concentrations of H⁺ and OH⁻). Kw increases with temperature — at 37°C (body temperature), Kw ≈ 2.4 × 10⁻¹⁴, and neutral pH ≈ 6.8.

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Water Autoionization Equilibrium

2H₂O ⇌ H₃O⁺ + OH⁻
or simplified: H₂O ⇌ H⁺ + OH⁻

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C

In pure water: [H⁺] = [OH⁻] = √(1.0 × 10⁻¹⁴) = 1.0 × 10⁻⁷ M.

pH and pOH

pH = −log₁₀[H⁺]. pOH = −log₁₀[OH⁻]. At 25°C: pH + pOH = pKw = 14.

In pure water at 25°C: pH = pOH = 7.0 → neutral. pH < 7: acidic. pH > 7: basic/alkaline.

Temperature Dependence

Kw increases with temperature (endothermic process). At 0°C: Kw ≈ 1.1 × 10⁻¹⁵; neutral pH ≈ 7.5. At 25°C: Kw = 1.0 × 10⁻¹⁴; neutral pH = 7.0. At 37°C: Kw ≈ 2.4 × 10⁻¹⁴; neutral pH ≈ 6.8. At 100°C: Kw ≈ 10⁻¹²; neutral pH = 6.0. Important: neutral pH is NOT always 7 — it equals pKw/2 at the relevant temperature.

Using Kw

Given [H⁺], calculate [OH⁻]: [OH⁻] = Kw/[H⁺]. Example: blood pH 7.4 → [H⁺] = 10⁻⁷·⁴ = 4.0 × 10⁻⁸ M; [OH⁻] = 10⁻¹⁴/4.0 × 10⁻⁸ = 2.5 × 10⁻⁷ M. Given a strong base: 0.01 M NaOH → [OH⁻] = 0.01 M → [H⁺] = 10⁻¹⁴/0.01 = 10⁻¹² M → pH = 12.

Glossary

Water Autoionization
H₂O ⇌ H⁺ + OH⁻; Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C; defines the pH scale; neutral pH = 7.0 at 25°C (pKw/2); endothermic — Kw increases with temperature.
Kw (Water Dissociation Constant)
The equilibrium constant for water autoionization: [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C; pKw = 14; neutral pH = pKw/2; increases with temperature.
pOH
−log₁₀[OH⁻]; related to pH by: pH + pOH = pKw = 14 (at 25°C); used to calculate [OH⁻] from pH and for strong base pH calculations.

Frequently Asked Questions

Water self-ionizes: H₂O ⇌ H⁺ + OH⁻ (or 2H₂O ⇌ H₃O⁺ + OH⁻). The equilibrium constant Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C. In pure water, [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M (because [H⁺] × [OH⁻] = 10⁻¹⁴ and they must be equal). This defines the neutral point. pKw = −log(Kw) = 14 at 25°C; neutral pH = pKw/2 = 7. Kw changes with temperature — the autoionization is endothermic, so Kw increases with heat.

Neutral pH = pKw/2. Since Kw is temperature-dependent, neutral pH changes too. At 25°C: Kw = 10⁻¹⁴; neutral pH = 7.0. At 37°C (body temperature): Kw ≈ 2.4 × 10⁻¹⁴; pKw = 13.62; neutral pH = 6.81. At 100°C: Kw ≈ 10⁻¹²; neutral pH = 6.0. Blood at 37°C has pH ≈ 7.35–7.45, which is slightly basic relative to neutral pH ≈ 6.8 at 37°C. A solution is neutral when [H⁺] = [OH⁻], regardless of the absolute pH value — and this depends on temperature.

Step 1: [H⁺] = 10^(−pH). Step 2: [OH⁻] = Kw/[H⁺] = 10⁻¹⁴/[H⁺]. Or directly: pOH = pKw − pH = 14 − pH (at 25°C); [OH⁻] = 10^(−pOH). Example: pH 4.5: [H⁺] = 10⁻⁴·⁵ = 3.16 × 10⁻⁵ M; [OH⁻] = 10⁻¹⁴/3.16 × 10⁻⁵ = 3.16 × 10⁻¹⁰ M; pOH = 14 − 4.5 = 9.5; [OH⁻] = 10⁻⁹·⁵ = 3.16 × 10⁻¹⁰ M ✓. pH 4.5 is acidic — [H⁺] > [OH⁻] by a factor of (3.16 × 10⁻⁵)/(3.16 × 10⁻¹⁰) = 10⁵ = 100,000.

For strong acids (HCl, HNO₃, H₂SO₄) and strong bases (NaOH, KOH): completely dissociated → [H⁺] or [OH⁻] = concentration. Strong acid: 0.05 M HCl → [H⁺] = 0.05 M; pH = −log(0.05) = 1.30. Strong base: 0.002 M NaOH → [OH⁻] = 0.002 M; [H⁺] = Kw/[OH⁻] = 10⁻¹⁴/0.002 = 5.0 × 10⁻¹² M; pH = −log(5.0 × 10⁻¹²) = 11.30. Alternatively: pOH = −log(0.002) = 2.70; pH = 14 − 2.70 = 11.30. For very dilute solutions (< 10⁻⁶ M acid), water autoionization contribution to [H⁺] cannot be ignored — use the quadratic equation.