All subjects · Chemistry · WCH14 (Unit 4)

The ionic product of water and the pH of bases

Water ionises very slightly. Its ionic product, Kw, links [H+] and [OH-] in every aqueous solution, which lets you calculate the pH of alkalis.

Kw (spec 14.10, 14.12)

H2O(l) ⇌ H+(aq) + OH-(aq)

Kw = [H+][OH-] = 1.00 x 10-14 mol2 dm-6 at 298 K; pKw = -log10 Kw = 14.00.

In pure water [H+] = [OH-], so [H+] = √Kw = 1.00 x 10-7 and pH = 7.00 at 298 K.

Temperature

The ionisation of water is endothermic, so Kw increases with temperature. At 50 °C pure water has pH below 7, but it is still neutral because [H+] still equals [OH-].

Strong bases (spec 14.11)

A strong base such as NaOH dissociates fully: [OH-] = concentration (twice the concentration for Ba(OH)2). Then

[H+] = Kw / [OH-], and pH = -log10[H+], or pH = pKw - pOH = 14.00 + log10[OH-].

Salts (spec 14.13)

Solutions of salts are not all neutral. Sodium ethanoate (weak acid, strong base) is slightly alkaline; ammonium chloride (strong acid, weak base) is slightly acidic; sodium chloride is neutral.

Worked example

Calculate the pH of 0.0200 mol dm-3 sodium hydroxide at 298 K.

  1. [OH-] = 0.0200 mol dm-3
  2. [H+] = 1.00 x 10-14 / 0.0200 = 5.00 x 10-13 mol dm-3
  3. pH = -log10(5.00 x 10-13)

Answer: pH = 12.30

Key idea: Kw = [H+][OH-] = 1.00 x 10-14 mol2 dm-6 at 298 K. For strong bases [H+] = Kw/[OH-]. Neutral means [H+] = [OH-], not always pH 7.

Check yourself

  1. Which expression is Kw?
    • [H+][OH-] / [H2O]
    • [H+] / [OH-]
    • [H+] + [OH-]
    • [H+][OH-]
    Show answer

    [H+][OH-]. [H2O] is effectively constant and is included in Kw.

  2. Calculate the pH of 0.0500 mol dm-3 potassium hydroxide at 298 K. (Kw = 1.00 x 10-14 mol2 dm-6)
    Show answer

    12.70. [H+] = 1.00 x 10-14 / 0.0500 = 2.00 x 10-13; pH = 12.70.

  3. Calculate the pH of 0.0500 mol dm-3 barium hydroxide, Ba(OH)2, at 298 K.
    Show answer

    13.00. [OH-] = 2 x 0.0500 = 0.100; [H+] = 1.00 x 10-13; pH = 13.00.

  4. At 50 °C, Kw = 5.48 x 10-14 mol2 dm-6. Calculate the pH of pure water at this temperature.
    Show answer

    6.63. [H+] = √(5.48 x 10-14) = 2.341 x 10-7; pH = 6.63.

  5. Pure water at 50 °C has a pH below 7. Why is it still neutral?
    • Kw has decreased
    • It contains no ions
    • [H+] still equals [OH-]
    • pH can only be measured at 25 °C
    Show answer

    [H+] still equals [OH-]. Neutral means equal [H+] and [OH-]; both increase as Kw increases.

  6. Which salt dissolves to give a solution with a pH above 7?
    • Sodium ethanoate
    • Sodium chloride
    • Ammonium chloride
    • Potassium nitrate
    Show answer

    Sodium ethanoate. Ethanoate ions accept protons from water, making OH-.

  7. What is pKw at 298 K?
    Show answer

    14.00. pKw = -log10(1.00 x 10-14) = 14.00.

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Last updated 7 November 2026. Spotted a mistake? Tell us.