2.5.3

The Ionic Product of Water (A2 Only)

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The Ionic Product of Water

Water is very slightly dissociated into its ions.

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Dissociation of water

  • Water very slightly dissociates into its ions according to the following equation:
    • 2H2O(l) ⇋ H3O+(aq) + OH-(aq)
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Equilibrium constant

  • The equilibrium constant for the dissociation of water is:
    • Kc = [H3O+][OH][H2O]\frac{[H_3O^+][OH^-]}{[H_2O]}
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Ionic product of water

  • The equilibrium constant can be simplified because the concentration of water is pretty much constant.
    • Water only dissociates slightly, so the concentration doesn’t really change.
    • We write a new equation: Kw = [H3O+(aq)][OH-(aq)]
    • It may also be written as: Kw = [H+(aq)][OH-(aq)]
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Temperature dependence of Kw

  • Kw is just an equilibrium constant.
    • Because of this, it behaves the same as any other equilibrium constant, Kc.
  • Importantly, this means it has a temperature dependence. The hotter the water is, the more dissociated it is and the larger its Kw value.

Use of Kw

Kw can be used to calculate the concentration of hydroxide ions given the pH.

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The value of Kw

  • At room temperature, it just so happens that Kw is around 1 ×10-14.
    • As the concentration of hydroxide ions must equal the concentration of hydrogen ions for pure water, we know the concentration of both must be 1 × 10-7.
    • If the [H+] is 1 ×10-7, then the pH is 7.
    • This is why we say a neutral solution has a pH of 7.
    • A neutral solution is one with the same concentration of hydroxide ions and hydrogen ions.
Illustrative background for Use of K<sub>w</sub>Illustrative background for Use of K<sub>w</sub> ?? "content

Use of Kw

  • The Kw is a constant at a specific temperature.
    • This means that at a specific temperature, we can relate the [H+] to the concentration of [OH].
  • If we know the pH of a solution, we can calculate the concentration of [OH].
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Example

  • Say we dissolve one mole of HCl into a litre of water:
    • The concentration of H+ ions is 1moldm-3.
    • We can rearrange the ion product equation from:
    • Kw = 1 × 10-14 = [H3O+(aq)][OH-(aq)]
  • To:
    • [OH-(aq)] = 1 × 10-14 ÷ 1moldm-3
      So we can calculate that [OH-(aq)] = 1 × 10-14.
    • The [OH-(aq)] is much lower in an acidic solution than in a neutral solution.

Jump to other topics

1Physical Chemistry

2Physical Chemistry 2 (A2 Only)

3Inorganic Chemistry

4Inorganic Chemistry 2 (A2 Only)

5Organic Chemistry 1

6Organic Chemistry 2 (A2 Only)

6.1Optical Isomerism (A2 Only)

6.2Aldehydes & Ketones (A2 Only)

6.3Carboxylic Acids & Esters (A2 Only)

6.4Aromatic Chemistry (A2 Only)

6.5Amines (A2 Only)

6.6Polymers (A2 Only)

6.7Biological Organic (A2 Only)

6.8Organic Synthesis (A2 Only)

6.9NMR Spectroscopy (A2 Only)

6.10Chromatography (A2 Only)

6.11A-A* (AO3/4) - Organic 2

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