Part 2: Weak acids, Ka and the limits of approximations

Derive the useful square-root relationship from an equilibrium balance and decide whether the assumptions are reasonable.

Ka measures equilibrium dissociation at a given temperature

A weak monobasic acid establishes HA(aq) ⇌ H⁺(aq) + A⁻(aq). Ka = [H⁺][A⁻]/[HA], using equilibrium concentrations. Water as solvent is absorbed into the constant. Within this convention Ka has units mol dm⁻³.

At the same temperature, a larger Ka means greater acid dissociation for comparable conditions and a stronger acid. pKa = −log₁₀Ka reverses the comparison: a lower pKa means a larger Ka. Diluting a given acid does not change its Ka at constant temperature, although its fraction ionised increases.

Why [H⁺] is approximately √(Ka c)

Let the initial acid concentration be c and let x mol dm⁻³ dissociate. With no added conjugate base and negligible contribution from water, equilibrium [H⁺] ≈ [A⁻] ≈ x and [HA] = c − x. Thus Ka ≈ x²/(c − x).

When dissociation is small relative to c, replace c − x by c to obtain [H⁺] ≈ √(Ka c). These are two distinct assumptions: water contributes negligible H⁺, and acid dissociation removes only a small fraction of HA. Neither assumption is automatic just because an acid is labelled weak.

Do not use this square-root expression for a buffer containing a substantial added amount of A⁻. In that mixture [A⁻] is not equal to [H⁺]; use the full Ka relationship with the supplied conjugate-pair amounts instead.

Worked example and a numerical assumption check

For original illustrative data 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. Therefore pH = 2.90.

The fraction dissociated is approximately 0.0012649/0.0800 = 0.0158, or 1.58%. This is small enough for a useful A-Level approximation. Keeping extra digits until the final logarithm avoids avoidable rounding drift. A five-percent comparison is a practical check, not an OCR rule that every question explicitly demands.

Now consider c = 0.0100 and Ka = 4.00 × 10⁻³ mol dm⁻³. The same shortcut predicts x = 0.00632, about 63% of c. Calling c − x approximately c is plainly unreasonable. Explain the breakdown rather than presenting the approximate pH as reliable. OCR does not require solving the resulting quadratic.

Worked example: calculate Ka from a measured pH

A freshly prepared monobasic acid has initial concentration 0.0500 mol dm⁻³ and measured pH 3.00. If water contribution is negligible, x = 10⁻³·⁰⁰ = 0.00100 mol dm⁻³. Equilibrium [HA] = 0.0490 mol dm⁻³, so Ka = 0.00100²/0.0490 = 2.04 × 10⁻⁵ mol dm⁻³.

Using the small-dissociation approximation would give 2.00 × 10⁻⁵. The close agreement is consistent with only two percent dissociation. If the question explicitly supplies equilibrium [HA], use it directly rather than subtracting x again.

An incomplete answer such as “the weak acid has fewer hydrogen ions” needs a concentration comparison and an explanation of partial dissociation. For equal initial molar concentrations, partial ionisation gives a lower [H⁺] and hence a higher pH than the corresponding strong monobasic acid.

Quick checks

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

Sources