Ion Product of Water Kw

Kw = [H+][OH−] = 1.0×10−14 at 25 °C ties pH and pOH together: pH + pOH = 14.00.

What it means

Water always autoionizes a little — H2O ⇌ H⁺ + OH⁻ — so every aqueous solution contains both ions, and their product is locked at Kw = 1.0 × 10⁻¹⁴ at 25 °C. That means knowing one ion's concentration instantly gives you the other: a strongly acidic solution's tiny [OH−] is simply Kw divided by its large [H⁺]. Taking negative logs gives the handy shortcut pH + pOH = 14.00, valid only at 25 °C since Kw itself changes with temperature.

Variables

Assumptions

• The solution is aqueous.
• The value Kw = 1.0 × 10⁻¹⁴ holds at 25 °C; Kw increases as temperature rises.
• Water's autoionization is at equilibrium.
• Concentrations approximate activities in these dilute solutions.

Derivation

Treat water's autoionization, H2O ⇌ H⁺ + OH⁻, like any equilibrium: Kw = [H⁺][OH⁻], with pure liquid water omitted as usual. Take the negative base-10 logarithm of both sides: −log Kw = −log[H⁺] + (−log[OH⁻]), giving pKw = pH + pOH. At 25 °C, pKw = −log(1.0 × 10⁻¹⁴) = 14.00.

Worked example

Given:

• pOH = 5.30
• Temperature 25 °C, so Kw = 1.0 × 10⁻¹⁴ and pH + pOH = 14.00

1. Use pH + pOH = 14.00: pH = 14.00 − 5.30 = 8.70.

2. [H⁺] = 10^(−pH) = 10^(−8.70).

3. Split the exponent: 10^(−8.70) = 10^0.30 × 10⁻⁹ = 1.995 × 10⁻⁹ M.

4. Round to three significant figures: [H⁺] = 2.00 × 10⁻⁹ M.

5. Check with Kw: [OH⁻] = 10^(−5.30) = 5.01 × 10⁻⁶ M, and (2.00 × 10⁻⁹)(5.01 × 10⁻⁶) = 1.00 × 10⁻¹⁴ = Kw.