Weak acids, Ka, Kw and strong basesEdexcel International A Level Chemistry: Revision notes

For a weak acid HA ⇌ H⁺ + A⁻, the acid dissociation constant is Ka = [H⁺][A⁻] / [HA] (units mol dm⁻³). A larger Ka means a stronger weak acid. Water is omitted because its concentration is effectively constant.

pKa = −log₁₀Ka, and Ka = 10⁻ᵖᴷᵃ. A smaller pKa means a stronger acid.

Example: for benzoic acid, Ka = 6.3 × 10⁻⁵ mol dm⁻³, so pKa = 4.20.

Two assumptions let you avoid a quadratic equation:

• [H⁺] = [A⁻], because the dissociation of the acid is the only significant source of H⁺ (the contribution from water is negligible)
• [HA]equilibrium ≈ [HA]initial, because only a tiny fraction of the acid dissociates

Then Ka = [H⁺]²/[HA], so [H⁺] = √(Ka × [HA]) and pH = −log₁₀[H⁺].

Worked example: 0.0100 mol dm⁻³ benzoic acid. [H⁺] = √(6.3 × 10⁻⁵ × 0.0100) = 7.94 × 10⁻⁴ mol dm⁻³, so pH = 3.10.

Water ionises slightly: H₂O ⇌ H⁺ + OH⁻. The ionic product of water is Kw = [H⁺][OH⁻], which is 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C. pKw = −log₁₀Kw = 14.00 at 25 °C.

For a strong base the [OH⁻] equals the base concentration (times two for Ba(OH)₂), then [H⁺] = Kw/[OH⁻]. For example, 0.0500 mol dm⁻³ NaOH gives [H⁺] = 2.00 × 10⁻¹³ mol dm⁻³ and pH = 12.70. Equivalently, pH = pKw − pOH.

The ionisation of water is endothermic, so Kw increases with temperature and pKw falls. At 50 °C, pKw = 13.26 and pure water has pH 6.63, but it is still neutral because [H⁺] = [OH⁻].

At equal concentration (for example 0.100 mol dm⁻³):

• Strong acid has a lower pH than a weak acid: HCl 1.00 and ethanoic acid 2.88
• Strong base has a higher pH than a weak base: NaOH 13.00 and ammonia 11.12
• Salts: NaCl is about 7; ammonium chloride is below 7 because NH₄⁺ donates protons to water; sodium ethanoate is above 7 because CH₃COO⁻ accepts protons from water

On dilution, a strong acid's pH rises by 1 for every tenfold dilution (10×, 100×, 1000× gives 1, 2, 3 units). A weak acid's pH rises by only about 0.5 per tenfold dilution, because more of it dissociates as it becomes more dilute.

To find Ka from experiment:

• Convert mass to amount: n = mass ÷ M, then divide by the volume in dm³ for [HA]
• Convert the measured pH to [H⁺] = 10⁻ᵖᴴ
• Apply the assumptions and use Ka = [H⁺]²/[HA]

Worked example: 0.610 g of benzoic acid (M = 122.0) in 100 cm³ gives pH 2.75. [HA] = (0.610 ÷ 122.0) ÷ 0.100 = 0.0500 mol dm⁻³; [H⁺] = 1.78 × 10⁻³; Ka = (1.78 × 10⁻³)² ÷ 0.0500 = 6.3 × 10⁻⁵ mol dm⁻³.

Measure the pH of a solution of known concentration of a weak acid with a calibrated pH meter. Calibrate with buffer solutions of known pH, rinse the probe with deionised water between solutions and stir gently before reading.

Use the pH to find [H⁺] and calculate Ka as above. Repeating with several concentrations, or by diluting the solution accurately with a volumetric flask and pipette, tests whether Ka is constant; a good set of results gives similar Ka values at each dilution.