Topic 14 · page 7 of 18

Water and strong bases

Water ionises very slightly, and the product of [H⁺] and [OH⁻] is the constant Kw. That constant turns the concentration of a strong base into a pH.

14 min read14.1014.1114.12

Read first: Strong and weak acids

The ionic product of water 14.10

Pure water is not just H₂O molecules. A very small number of them transfer a proton to a neighbour:

HX2O(l)⇌HX+(aq)+OHX−(aq) \ce{H2O(l) <=> H+(aq) + OH-(aq)}

This is an equilibrium, so it has an equilibrium constant. Water is the solvent and its concentration is effectively constant, so it is left out and the constant is written with the two ions only.

The units are (mol dm⁻³)² = mol² dm⁻⁶. The expression holds for any aqueous solution, not just pure water. If an acid raises [H⁺], then [OH⁻] must fall so that the product stays at KwK_\text{w}. If a base raises [OH⁻], [H⁺] must fall.

In pure water each ionisation makes one H⁺ and one OH⁻, so [H⁺] = [OH⁻]. At 298 K:

[HX+]=1.00×10−14=1.00×10−7 mol dm−3pH=7.00 [\ce{H+}] = \sqrt{1.00\times10^{-14}} = 1.00\times10^{-7}\ \text{mol dm}^{-3} \qquad \text{pH} = 7.00

pKw 14.12

The same trick as pH and pKa\text{p}K_\text{a} makes KwK_\text{w} easier to handle.

Take logs of Kw=[HX+][OHX−]K_\text{w} = [\ce{H+}][\ce{OH-}] and change the signs. It becomes a sum:

pKw=pH+pOH \text{p}K_\text{w} = \text{pH} + \text{pOH}

where pOH = −lg [OH⁻]. This is a convenient shortcut for calculating pH from [OH⁻], but you can always avoid pOH by using KwK_\text{w} directly.

The pH of a strong base 14.11

A strong base is fully dissociated, so [OH⁻] is fixed by the concentration and the number of OH⁻ in the formula.

[OHX−]=(number of OH− per formula unit)×c(base) [\ce{OH-}] = (\text{number of OH}^-\text{ per formula unit}) \times c(\text{base})

Then use KwK_\text{w} to find [H⁺], and take the log.

Worked exampleSodium hydroxide, using Kw

Calculate the pH of 0.020 mol dm⁻³ NaOH at 298 K.

  1. NaOH is a strong base: [OH⁻] = 0.020 mol dm⁻³.
  2. [H⁺] = KwK_\text{w} ÷ [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 0.020 = 5.0 × 10⁻¹³ mol dm⁻³
  3. pH = −lg (5.0 × 10⁻¹³) = 12.30

Worked exampleThe same answer, using pKw

  1. pOH = −lg (0.020) = 1.70
  2. pH = pKw\text{p}K_\text{w} − pOH = 14.00 − 1.70 = 12.30

Worked exampleBarium hydroxide

Calculate the pH of 0.0050 mol dm⁻³ Ba(OH)₂ at 298 K.

  1. Each formula unit releases two OH⁻: [OH⁻] = 2 × 0.0050 = 0.010 mol dm⁻³.
  2. [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.010 = 1.0 × 10⁻¹² mol dm⁻³
  3. pH = 12.00

Kw changes with temperature 14.1014.11

The ionisation of water is endothermic, so raising the temperature shifts the equilibrium to the right and KwK_\text{w} rises. Use the value for the temperature in the question.

Temperature / °C KwK_\text{w} / mol² dm⁻⁶ pKw\text{p}K_\text{w} pH of pure water
25 1.00 × 10⁻¹⁴ 14.00 7.00
50 5.5 × 10⁻¹⁴ 13.26 6.63
100 5.1 × 10⁻¹³ 12.29 6.15

Worked exampleSodium hydroxide at 50 °C

Calculate the pH of 0.020 mol dm⁻³ NaOH at 50 °C, where KwK_\text{w} = 5.5 × 10⁻¹⁴ mol² dm⁻⁶.

  1. [OH⁻] = 0.020 mol dm⁻³
  2. [H⁺] = 5.5 × 10⁻¹⁴ ÷ 0.020 = 2.75 × 10⁻¹² mol dm⁻³
  3. pH = −lg (2.75 × 10⁻¹²) = 11.56

At 298 K the same solution had pH 12.30. The pH of an alkali falls as the temperature rises, but the solution is not any less alkaline: [OH⁻] has not changed.

Worked examplePure water at 60 °C

At 60 °C, KwK_\text{w} = 9.6 × 10⁻¹⁴ mol² dm⁻⁶. Calculate the pH of pure water and say whether it is neutral.

  1. [H⁺] = [OH⁻], so [H⁺]² = 9.6 × 10⁻¹⁴ and [H⁺] = 3.10 × 10⁻⁷ mol dm⁻³.
  2. pH = −lg (3.10 × 10⁻⁷) = 6.51
  3. The water is still neutral, because [H⁺] = [OH⁻].

Check yourself

  1. Define the ionic product of water and state its value and units at 298 K.
  2. Calculate the pH of 0.040 mol dm⁻³ NaOH at 298 K.
  3. Calculate the pH of 0.0150 mol dm⁻³ Ba(OH)₂ at 298 K.
  4. A solution at 298 K has pH 9.40. Calculate [OH⁻].
  5. At 100 °C, KwK_\text{w} = 5.1 × 10⁻¹³ mol² dm⁻⁶. Calculate the pH of 0.100 mol dm⁻³ NaOH at this temperature.
  6. At 50 °C, KwK_\text{w} = 5.5 × 10⁻¹⁴ mol² dm⁻⁶. Calculate the pH of pure water at 50 °C and explain why it is still neutral.

Answers cover these until you have tried the questions

  1. KwK_\text{w} is the product of the equilibrium concentrations of H⁺ and OH⁻ in water: Kw=[HX+][OHX−]K_\text{w} = [\ce{H+}][\ce{OH-}]. At 298 K it is 1.00 × 10⁻¹⁴ mol² dm⁻⁶.
  2. [OH⁻] = 0.040. [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.040 = 2.5 × 10⁻¹³ mol dm⁻³. pH = −lg (2.5 × 10⁻¹³) = 12.60
  3. [OH⁻] = 2 × 0.0150 = 0.0300 mol dm⁻³. [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0300 = 3.33 × 10⁻¹³ mol dm⁻³. pH = 12.48
  4. [H⁺] = 10−9.4010^{-9.40} = 3.98 × 10⁻¹⁰ mol dm⁻³. [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 3.98 × 10⁻¹⁰ = 2.5 × 10⁻⁵ mol dm⁻³
  5. [OH⁻] = 0.100. [H⁺] = 5.1 × 10⁻¹³ ÷ 0.100 = 5.1 × 10⁻¹² mol dm⁻³. pH = −lg (5.1 × 10⁻¹²) = 11.29
  6. [H⁺] = [OH⁻] = √(5.5 × 10⁻¹⁴) = 2.35 × 10⁻⁷ mol dm⁻³. pH = 6.63. It is still neutral because [H⁺] = [OH⁻], even though the pH is not 7.