Part 3: Weak acids, Ka and pKa

Derive the weak-acid approximation, calculate pH and check when the approximation is reasonable.

Write the equilibrium before the expression

For a monoprotic weak acid HA, Ka relates the equilibrium concentrations of hydrogen ions, conjugate base and undissociated acid. Water is incorporated into the constant in the dilute aqueous treatment. A larger Ka means greater dissociation and a stronger acid for a comparable acid equilibrium at the same temperature.

In the A-level concentration convention Ka has units mol dm⁻³. pKa = −log₁₀Ka, so smaller pKa means stronger acid. Neither is a direct measure of how much acid was initially dissolved.

Understand the two approximations

For an acid-only solution with initial concentration c and no appreciable added conjugate base or other acid, let x be the amount dissociated per dm³. Neglecting water’s ion contribution gives [H⁺] = [A⁻] = x and [HA] = c − x. Therefore Ka = x²/(c − x).

If dissociation is small compared with c, replace c − x by c. Then [H⁺] ≈ √(Kac). This is not the buffer formula. It fails when added salt means [A⁻] is much larger than [H⁺], and the small-dissociation assumption can fail for a very dilute or relatively strongly dissociating acid.

Worked pH and percentage dissociation

For an original practice acid with c = 0.0800 mol dm⁻³ and Ka = 2.00 × 10⁻⁵ mol dm⁻³, [H⁺] ≈ √(1.60 × 10⁻⁶) = 1.2649 × 10⁻³ mol dm⁻³. pH = 2.90.

Percentage dissociation ≈ 100x/c = 1.58%, so replacing c − x by c is reasonable for an ordinary A-level estimate. As a practical check, a few per cent dissociation often supports the approximation, but required accuracy governs the decision; “5%” is a working guideline rather than a universal exam rule.

If the approximation is unsuitable but water remains negligible, solve x² + Kax − Kac = 0 and take the positive root. With the values above, the more exact x is 1.2550 × 10⁻³ mol dm⁻³, still giving pH 2.90 to two decimal places.

Find Ka or concentration from pH

For an acid-only solution with c = 0.0500 mol dm⁻³ and pH = 3.00, x = 1.00 × 10⁻³ mol dm⁻³. Using the small-dissociation approximation, Ka ≈ x²/c = 2.00 × 10⁻⁵ mol dm⁻³. Keeping c − x gives 2.04 × 10⁻⁵ mol dm⁻³. State which treatment the question asks for.

Rearranging the approximate equation gives c ≈ [H⁺]²/Ka. For a practice acid with pH = 3.20 and Ka = 1.50 × 10⁻⁵, c ≈ 0.0265 mol dm⁻³. The corresponding pKa is 4.82. Check that the resulting [H⁺]/c is small before trusting the approximation.

Dilution changes dissociation as well as concentration

At fixed temperature, dilution does not change Ka. For an acid-only weak-acid solution within the square-root approximation, a tenfold decrease in c decreases [H⁺] by √10 and raises pH by about 0.50. The fraction dissociated increases because x/c ≈ √(Ka/c).

This differs from the approximately one-unit pH rise for tenfold dilution of a strong monoprotic acid. Neither simple rule applies without limit as the concentration approaches water’s own ion contribution. Weak bases such as ammonia also establish equilibria with water, but this specification’s quantitative weak-acid work centres on Ka.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources