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Acid-Base Equilibria

Water, Kw and the pH scale

  • 3 things to know
  • A2 Unit 3

Overview

Recall questions on Water and the pH scale.

What you need to know

Answer each one before you reveal it.

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Key term

Write the expression for the ionic product of water and state its value at 298 K.

Key term

State the units of Kw.

Numbers

At 310 K, Kw = 2.4 × 10⁻¹⁴ mol² dm⁻⁶. Calculate the pH of pure water at body temperature.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Write the equation for the ionisation of water and explain why it happens at all.

    Model answer

    H₂O(l) ⇌ H⁺(aq) + OH⁻(aq), or written as a proton transfer between two molecules, 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq). Water is very slightly amphoteric, so one molecule can pass a proton to another, although at any moment only about one molecule in 5.6 × 10⁸ has done so.

  2. 2

    Write the expression for the ionic product of water and state its value at 298 K.

    Model answer

    Kw = [H⁺][OH⁻], and at 298 K its value is 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

  3. 3

    Why does [H₂O] not appear in the expression for Kw?

    Model answer

    Only a minute fraction of the water ionises, so the concentration of undissociated water stays effectively constant at about 55.5 mol dm⁻³. That constant is combined with the equilibrium constant to give the simpler quantity Kw.

  4. 4

    State the units of Kw.

    Model answer

    The units are mol² dm⁻⁶, because Kw is the product of two concentrations each measured in mol dm⁻³.

  5. 5

    How is pH defined?

    Model answer

    pH = −log₁₀[H⁺], where [H⁺] is the hydrogen ion concentration in mol dm⁻³. To go the other way, [H⁺] = 10⁻ᵖᴴ.

  6. 6

    Show that pH + pOH = 14.00 at 298 K.

    Model answer

    Since [H⁺][OH⁻] = 1.00 × 10⁻¹⁴, taking negative logarithms of both sides gives −log[H⁺] − log[OH⁻] = −log(1.00 × 10⁻¹⁴). That is pH + pOH = 14.00, so pKw = 14.00 at this temperature.

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