pH calculations and buffers

pH measures the concentration of hydrogen ions: pH = −log[H⁺]. A strong acid is fully dissociated, so [H⁺] is the concentration of the acid. A weak acid only partly dissociates, and its acid dissociation constant, Ka, says how far. Water has its own ionic product, Kw, which links [H⁺] and [OH⁻]. A buffer is a mixture that keeps the pH almost the same when a little acid or alkali is added.

Key facts

  • pH = −log[H⁺] and [H⁺] = 10^(−pH). Each unit of pH is a factor of 10 in [H⁺].
  • Strong acids such as HCl and HNO₃ are fully dissociated: [H⁺] equals the concentration of the acid.
  • Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. For a strong base find [OH⁻], then [H⁺] = Kw ÷ [OH⁻]. Kw grows with temperature, so pure water at 50 °C has a pH of 6.63, but it is still neutral.
  • For a weak acid, HA ⇌ H⁺ + A⁻: Ka = [H⁺][A⁻] ÷ [HA], and pKa = −log Ka. The larger Ka (the smaller pKa), the stronger the acid.
  • pH of a weak acid: take [H⁺] = [A⁻] and assume [HA] stays at its starting value, so [H⁺] = √(Ka × [HA]).
  • A buffer resists changes in pH. An acidic buffer is a weak acid with its salt, such as ethanoic acid with sodium ethanoate: [H⁺] = Ka × [HA] ÷ [A⁻].
  • At half-neutralisation of a weak acid by a strong base, [HA] = [A⁻], so pH = pKa.

Worked examples

Example 1

Calculate the pH of 0.025 mol dm⁻³ hydrochloric acid and of 0.025 mol dm⁻³ sodium hydroxide at 298 K.

  1. Hydrochloric acid is fully dissociated. [H⁺] = 0.025; pH = −log 0.025 = 1.60
  2. Sodium hydroxide is fully dissociated. [OH⁻] = 0.025
  3. Use the ionic product of water. [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.025 = 4.00 × 10⁻¹³
  4. Take the log. pH = −log(4.00 × 10⁻¹³) = 12.40

Answer: HCl: pH 1.60; NaOH: pH 12.40

Example 2

Calculate the pH of 0.100 mol dm⁻³ ethanoic acid. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

  1. Write Ka. Each molecule that dissociates gives one H⁺ and one A⁻, so [H⁺] = [A⁻]. Ka = [H⁺]² ÷ [HA]
  2. The acid is weak, so [HA] is still about 0.100. [H⁺]² = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶
  3. Take the square root. [H⁺] = 1.32 × 10⁻³ mol dm⁻³
  4. Take the log. pH = −log(1.32 × 10⁻³) = 2.88

Answer: pH = 2.88

Example 3

A buffer contains 0.10 mol dm⁻³ ethanoic acid and 0.20 mol dm⁻³ sodium ethanoate. Ka = 1.74 × 10⁻⁵ mol dm⁻³. Calculate its pH.

  1. In a buffer, [HA] is the acid’s concentration and [A⁻] is the salt’s. [H⁺] = Ka × [HA] ÷ [A⁻]
  2. Put in the numbers. [H⁺] = 1.74 × 10⁻⁵ × 0.10 ÷ 0.20 = 8.70 × 10⁻⁶
  3. Take the log. pH = −log(8.70 × 10⁻⁶) = 5.06

Answer: pH = 5.06