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OCR A Chemistry H432 · Year 13 · 5.1.3

Part 3: Buffer action, preparation and pH calculations

All 4 parts available. Reviewed 6 October 2026.

Explain which component removes added acid or alkali, then calculate buffer composition after any neutralisation has occurred.

Two components provide two different responses

A buffer minimises pH change when small amounts of acid or base are added. An acid buffer contains appreciable amounts of a weak acid HA and its conjugate base A⁻, commonly supplied by a soluble salt. The weak acid is a reservoir that removes added OH⁻; the conjugate base removes added H⁺.

Added H⁺ reacts with A⁻ to form HA. Added OH⁻ reacts with HA to form A⁻ and water. In equilibrium language, removing H⁺ causes more HA to dissociate. The ratio [HA]/[A⁻] changes only slightly when the added amount is small compared with both reservoirs, so [H⁺] changes only slightly.

A buffer does not hold pH perfectly constant and cannot absorb unlimited acid or alkali. If one component is nearly exhausted, its relevant protective response is lost. Dilution approximately preserves the ratio and therefore pH, but reduces capacity per unit volume; extreme dilution also undermines the usual approximations.

A⁻ + H⁺ → HA
HA + OH⁻ → A⁻ + H₂O
[H⁺] ≈ Ka × [HA]/[A⁻]

Worked example: mix the acid and its salt

For an original illustrative buffer, [HA] = 0.120 and [A⁻] = 0.180 mol dm⁻³ with Ka = 1.80 × 10⁻⁵ mol dm⁻³. Rearrangement gives [H⁺] = 1.80 × 10⁻⁵ × 0.120/0.180 = 1.20 × 10⁻⁵ mol dm⁻³ and pH = 4.92.

The same result follows from pH = pKa + log₁₀([A⁻]/[HA]), which is simply the logarithmic rearrangement. Starting from Ka reduces the risk of reversing the ratio. If both species occupy the same final volume, their mole ratio equals their concentration ratio. Volumes only cancel inside that ratio, not in every part of a calculation.

To prepare a specified buffer, calculate the required ratio, then choose actual acid and salt amounts and make up to the required volume using suitable volumetric apparatus. A target pH fixes a ratio, not a unique total concentration or capacity.

Worked example: create a buffer with a strong alkali

Mix 50.0 cm³ of 0.200 mol dm⁻³ HA with 20.0 cm³ of 0.200 mol dm⁻³ NaOH. Initial HA is 0.0100 mol and OH⁻ is 0.00400 mol. The reaction HA + OH⁻ → A⁻ + H₂O first leaves 0.00600 mol HA and forms 0.00400 mol A⁻. There is no appreciable excess strong base.

With Ka = 1.80 × 10⁻⁵, [H⁺] ≈ Ka × 0.00600/0.00400 = 2.70 × 10⁻⁵ mol dm⁻³, giving pH = 4.57. Both concentrations contain the same final-volume divisor, which cancels. Substituting the original 0.0100 mol HA would ignore neutralisation.

If 0.00100 mol strong acid is now added with negligible volume change, it consumes A⁻ and produces HA: new amounts are 0.00300 and 0.00700 mol respectively. Then [H⁺] = 4.20 × 10⁻⁵ and pH = 4.38. The pH decreases, but both reservoirs remain. This is a changed-condition calculation, not merely a repeated formula substitution.

Apply the conjugate-pair idea to blood plasma

The H₂CO₃/HCO₃⁻ pair helps maintain blood-plasma pH within about 7.35–7.45. Added acid is taken up by HCO₃⁻; added base is neutralised by H₂CO₃. Write the corresponding equations to show the chemistry rather than simply saying “the buffer absorbs acid”.

This buffer operates within a physiological system that also controls carbon dioxide and ions. It is not an isolated beaker, and the equilibrium model alone is not medical guidance. For the chemistry specification, explain the complementary roles of carbonic acid and hydrogencarbonate.

HCO₃⁻ + H⁺ → H₂CO₃
H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O

Build a connected buffer explanation

Name the pair, identify the added ion, show the reacting component and connect the small ratio change to the small [H⁺] and pH change. In a calculation, label initial amounts, neutralisation changes and final buffer amounts. A numerical answer without a species label is much harder to audit.

For an original extended response, explain how to prepare the partial-neutralisation buffer above and how it responds to added OH⁻. A good indicative response combines the mole balance, Ka calculation and HA + OH⁻ reaction. It does not assign a fictional official mark to every sentence.

Quick checks

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

Q1. Why is ethanoic acid alone a much less effective buffer against added strong acid than ethanoic acid with sodium ethanoate?Show answer

The acid alone contains only a small concentration of ethanoate from dissociation. Added salt supplies a substantial conjugate-base reservoir to remove H⁺. Effective buffering in both directions requires appreciable amounts of both components.

Q2. For Ka = 2.00 × 10⁻⁵ and n(HA):n(A⁻) = 1:4, find pH.Show answer

[H⁺] = 2.00 × 10⁻⁵/4 = 5.00 × 10⁻⁶ mol dm⁻³. pH = 5.30. The mole ratio can be used because both species share the same solution volume.

Q3. 0.0120 mol HA is mixed with 0.00300 mol OH⁻. With Ka = 2.00 × 10⁻⁵, estimate pH.Show answer

Neutralisation leaves 0.00900 mol HA and forms 0.00300 mol A⁻. [H⁺] = 2.00 × 10⁻⁵ × 3 = 6.00 × 10⁻⁵ mol dm⁻³. pH = 4.22.

Q4. A buffer has 0.0200 mol HA and 0.0100 mol A⁻. What amounts remain after adding 0.00200 mol OH⁻?Show answer

OH⁻ removes 0.00200 mol HA and produces that amount of A⁻. The new amounts are HA 0.0180 and A⁻ 0.0120 mol. The ratio falls from 2.00 to 1.50, so [H⁺] falls and pH rises.

Q5. Why can two buffers have the same pH but different capacity?Show answer

The same HA:A⁻ ratio gives approximately the same pH at the same Ka. Larger absolute amounts can neutralise more added acid or alkali before either reservoir becomes limiting. Ratio determines approximate pH; inventory affects capacity.

Sources

Sources and examiner guidance (reviewed 6 October 2026)

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