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5.3.2Physical Chemistry and Transition Elements

Ka, pH and Kw

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Section 1 of 5The Acid Dissociation Constant

5.3.2 Ka, pH and Kw

For a weak acid HA ⇌ H⁺ + A⁻ the equilibrium constant describes how far it ionises.

Ka expression
  • Ka = [H⁺][A⁻] / [HA], with units mol dm⁻³, because water is in large excess and is omitted.
Meaning
  • A larger Ka means more dissociation and a stronger acid, while a smaller Ka shows a weaker acid.
pKa
  • pKa = −log₁₀Ka, so a stronger acid has a smaller pKa, and Ka = 10−pKa.
Temperature
  • Ka is constant at a given temperature, and only a change of temperature alters its value.

Section 2 of 5pH and the Ionic Product of Water

Water ionises very slightly and this sets the scale for the pH range.

1
The pH is defined as pH = −log₁₀[H⁺], and [H⁺] = 10−pH, with [H⁺] in mol dm⁻³.
2
Water ionises as H₂O ⇌ H⁺ + OH⁻, and its equilibrium constant is combined with the constant [H₂O] to give Kw = [H⁺][OH⁻].
3
At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶, so in pure water [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³ and pH = 7.
4
For a strong base, [OH⁻] equals the base concentration for a monoacidic base, and [H⁺] = Kw ÷ [OH⁻].
5
Ionisation of water is endothermic, so Kw increases with temperature, and neutral water then has a pH below 7 while still [H⁺] = [OH⁻].

Section 3 of 5Strong and Weak Acids Compared

Strong and Weak Acids Compared
Strong acid
  • Fully dissociated, so [H⁺] equals the acid concentration, and 0.010 mol dm⁻³ HCl has pH 2.00.
Weak acid
  • Partially dissociated, so [H⁺] is much less than the acid concentration, and the pH is higher than for a strong acid of equal concentration.
Strong base
  • Fully dissociated, so NaOH of 0.10 mol dm⁻³ gives [OH⁻] = 0.10 mol dm⁻³ and pH 13 at 298 K.
Neutral solution
  • A solution is neutral when [H⁺] = [OH⁻], which is not always at pH 7 unless the temperature is 298 K.

Section 4 of 5Worked Example

Question

  • (a) Calculate the pKa of ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³.
  • (b) Calculate the pH of 0.050 mol dm⁻³ NaOH at 298 K.

  1. pKa = −log(1.74 × 10⁻⁵) = 4.76.
  2. NaOH is a strong base, so [OH⁻] = 0.050 mol dm⁻³.
  3. [H⁺] = Kw ÷ [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 0.050 = 2.0 × 10⁻¹³ mol dm⁻³.
  4. pH = −log(2.0 × 10⁻¹³) = 12.70.

Answer (a) pKa = 4.76; (b) pH = 12.70.

Section 5 of 5Exam Checklist & Tips

5.3.2 – Key Points for the Exam

  • Ka The acid dissociation constant, [H⁺][A⁻]/[HA] for a weak acid.
  • pKa The negative logarithm to base 10 of Ka.
  • pH The negative logarithm to base 10 of the hydrogen ion concentration.
  • Kw The ionic product of water, [H⁺][OH⁻], equal to 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K.

Common Mistakes

  • Using the pH of a weak acid as if it equalled −log of the acid concentration.
  • Forgetting that Kw changes with temperature and treating pH 7 as neutral at every temperature.
  • Quoting pH to the wrong number of decimal places;
    • the decimal places should match the significant figures of the concentration.
  • Stating that a larger pKa means a stronger acid, when it means a weaker acid.
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