Part 2: Weak-acid equilibria, Ka and dilution

Choose the weak-acid model, test its assumptions and turn a known mass and a measured pH into a dissociation constant.

Ka measures a particular acid equilibrium

For HA(aq) ⇌ H⁺(aq) + A⁻(aq), Ka = [H⁺][A⁻]/[HA], with all concentrations at equilibrium. Ka has units mol dm⁻³ in this convention. Water is omitted because it is the effectively constant solvent. A larger Ka means more extensive proton donation for otherwise comparable conditions; a smaller pKa means a stronger acid.

For ethanoic acid, Ka = [H⁺][CH₃COO⁻]/[CH₃COOH]. For NH₄⁺ acting as an acid, Ka = [H⁺][NH₃]/[NH₄⁺]. The acid need not be a neutral molecule; write its dissociation equation to identify numerator and denominator.

Ka is a constant for a stated acid and temperature. Dilution changes the degree of dissociation and pH, but not Ka at the same temperature. Do not describe Ka as the concentration of acid or assume a measured pH is itself pKa.

Worked example with an approximation check

A monoprotic weak acid has analytical concentration c = 0.0800 mol dm⁻³ and Ka = 1.80 × 10⁻⁵ mol dm⁻³. Let x be the concentration that dissociates. With negligible H⁺ from water and no added conjugate base, [H⁺] ≈ [A⁻] = x and [HA] = c − x. Hence Ka = x²/(c − x).

If dissociation is small, c − x ≈ c and x ≈ √(Ka c) = √(1.80 × 10⁻⁵ × 0.0800) = 1.20 × 10⁻³ mol dm⁻³. Then pH = 2.92. The estimated fraction dissociated is 0.00120/0.0800 = 0.0150, or 1.50%, so the approximation is reasonable.

The two assumptions are distinct: [A⁻] ≈ [H⁺] requires no substantial other source of either ion, and [HA] ≈ c requires small dissociation. A buffer violates the first assumption because salt deliberately supplies A⁻. A dilute or relatively strong weak acid may violate the second.

If needed, solve x² + Ka x − Ka c = 0 using the positive root x = [−Ka + √(Ka² + 4Ka c)]/2. This is an explanatory extension for testing the approximation, not a replacement for stating the assumptions requested in a question. If x becomes comparable to the water ion concentration, water must also be included.

Worked example: known mass and measured pH

An original experimental example dissolves 1.20 g of a monoprotic acid of molar mass 60.0 g mol⁻¹ and makes the solution up to 250.0 cm³. Its measured pH is 2.93. First calculate the analytical concentration: n = 1.20/60.0 = 0.0200 mol; c = 0.0200/0.2500 = 0.0800 mol dm⁻³.

[H⁺] = 10⁻²·⁹³ = 1.1749 × 10⁻³ mol dm⁻³. Taking [A⁻] ≈ [H⁺] and retaining acid lost by dissociation gives [HA] = 0.0800 − 0.0011749 = 0.0788251 mol dm⁻³. Ka = (1.1749 × 10⁻³)²/0.0788251 = 1.75 × 10⁻⁵ mol dm⁻³.

Using [HA] ≈ 0.0800 would instead give 1.73 × 10⁻⁵ mol dm⁻³, consistent with the small-dissociation approximation. State which calculation is used. The precision is limited by the pH measurement as well as the mass, flask and molar mass.

A pH reading 0.05 units too high multiplies the inferred [H⁺] by 10⁻⁰·⁰⁵ = 0.891. In the approximation Ka = [H⁺]²/c, this makes Ka about 0.794 of its correct value, roughly 21% too low. Repeating the same miscalibrated probe does not remove this systematic error.

Dilution increases the fraction dissociated

Diluting a strong monoprotic acid tenfold reduces [H⁺] approximately tenfold and raises pH by 1.00, while the acid dominates water. For a weak acid in the small-dissociation range, [H⁺] ≈ √(Ka c): a tenfold reduction of c reduces [H⁺] by √10 and raises pH by about 0.50.

The weak-acid equilibrium responds to dilution by increasing the fraction of acid molecules dissociated. Nevertheless [H⁺] decreases because dilution spreads the ions through a larger volume. “More dissociation” does not mean a greater hydrogen-ion concentration than before dilution.

The illustrative table compares an initially 0.100 mol dm⁻³ strong acid and a weak acid with Ka = 1.80 × 10⁻⁵ mol dm⁻³. Weak-acid values use x²/(c − x), with water neglected. After 1000-fold dilution the weak acid is about 34% dissociated, so the simple 0.50-per-tenfold rule is no longer accurate.

Why weak acids still react completely with alkali

The net neutralisation of a strong acid and strong alkali is H⁺(aq) + OH⁻(aq) → H₂O(l). Similar dilute strong acid–alkali pairs have similar molar enthalpies because the same net ionic process occurs.

For a weak acid, HA must also dissociate as OH⁻ removes H⁺. If dissociation absorbs energy, as in the standard school examples, the overall neutralisation is less exothermic: the heat released on forming water is partly offset by ionisation. The enthalpy effect depends on the actual acid; weakness alone is not a universal numerical enthalpy rule.

The standard enthalpy change of neutralisation refers to formation of one mole of water under the stated standard conditions. If 0.0200 mol water is formed and the reacting system releases 1.06 kJ, ΔneutH = −1.06/0.0200 = −53.0 kJ mol⁻¹. Use water formed or reaction extent, not the sum of acid and base moles.

Pearson 9CH0/01 June 2023 Q6(d)(iii) explicitly distinguished partial dissociation from neutralising capacity. Equal moles of monobasic weak and strong acids require equal moles of NaOH for completion; fewer initially free H⁺ ions do not imply a smaller total titre.

Quick checks

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

Sources