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Buffer solutionsEdexcel International A Level Chemistry: Revision notes

Section 1

What a buffer is and how it works

A buffer solution is one that resists changes in pH when small amounts of acid or alkali are added. An acidic buffer contains a weak acid (HA) and its conjugate base (A⁻), usually as a salt, for example ethanoic acid with sodium ethanoate.

The weak acid is only slightly dissociated, but the salt supplies plenty of A⁻, so both are present in large amounts:

  • Added acid: H⁺ + A⁻ → HA. The conjugate base removes the H⁺.
  • Added alkali: OH⁻ + HA → A⁻ + H₂O. The weak acid removes the OH⁻.

The ratio [HA]/[A⁻] changes only slightly, so [H⁺], and hence pH, stays almost constant.

Key termsbuffer solutionconjugate base
Common mistake

Explaining buffer action as the buffer 'cancelling' the acid. Name the species that reacts and give the equation.

Section 2

Calculating the pH of a buffer

From Ka = [H⁺][A⁻]/[HA], rearrange for [H⁺]:

[H⁺] = Ka × [HA]/[A⁻]

Then pH = −log₁₀[H⁺]. Because the volume is the same for both species, you can use amounts of HA and A⁻ in place of concentrations.

Worked example: ethanoic acid 0.200 mol dm⁻³, sodium ethanoate 0.100 mol dm⁻³, Ka = 1.74 × 10⁻⁵ mol dm⁻³. [H⁺] = 1.74 × 10⁻⁵ × 0.200/0.100 = 3.48 × 10⁻⁵ mol dm⁻³ so pH = 4.46.

If [HA] = [A⁻], then [H⁺] = Ka and pH = pKa.

Key termspH = pKa
Exam tip

Check direction: if there is more acid than base, the pH should be below the pKa.

Section 3

Preparing a buffer of a given pH

To make a buffer of a required pH, find [H⁺] = 10⁻ᵖᴴ, then use the rearranged expression to get the ratio [A⁻]/[HA] = Ka/[H⁺].

Worked example: pH 5.00 with ethanoic acid at 0.500 mol dm⁻³ in 500 cm³. [H⁺] = 1.00 × 10⁻⁵, so [A⁻]/[HA] = 1.74 × 10⁻⁵ ÷ 1.00 × 10⁻⁵ = 1.74. [A⁻] = 0.870 mol dm⁻³, which is 0.435 mol in 500 cm³, so the mass of CH₃COONa (M = 82.0) is 35.7 g.

Adding a small amount of acid changes the amounts, not the pH much: adding 5.0 × 10⁻³ mol H⁺ gives HA 0.255 mol and A⁻ 0.430 mol, pH 4.99, a change of only 0.01.

Common mistake

Using the amounts before reaction when strong acid or alkali is added. Subtract from A⁻ (or HA) first, then use the new ratio.

Section 4

Buffer action on a titration curve

When a weak acid is titrated with a strong base (or a weak base with a strong acid), the region before equivalence is a buffer: the curve rises slowly and flattens.

At the half-neutralisation point (half the volume needed for equivalence), half of the acid has been converted to A⁻, so [HA] = [A⁻] and pH = pKa. Hence Ka can be found from the pH at this volume: if the pH is 4.20, Ka = 10⁻⁴·²⁰ = 6.31 × 10⁻⁵ mol dm⁻³.

The buffer is most effective when [HA] and [A⁻] are similar, around pH = pKa.

Key termshalf-neutralisation point
Exam tip

Read the equivalence volume first, halve it, then read the pH at that volume from the curve.

Section 5

Why buffers matter in biology and food

Enzymes work only in a narrow pH range, so cells and body fluids must keep a nearly constant pH. In blood (pH about 7.4) the buffer is H₂CO₃/HCO₃⁻:

  • Added H⁺ is removed: HCO₃⁻ + H⁺ → H₂CO₃
  • Added OH⁻ is removed: H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O

In foods, buffers prevent deterioration: as bacteria and fungi grow they produce acids or alkalis, and a buffer resists the resulting pH change, keeping the food stable.

Key termsH₂CO₃/HCO₃⁻ buffer
Exam tip

Link the pH to the enzyme: a pH change alters the shape of the active site and the enzyme denatures.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Buffer solutions

  1. A technician prepares solutions containing ethanoic acid and sodium ethanoate. The Ka of ethanoic acid is 1.74 × 10⁻⁵ mol dm⁻³, so its pKa is 4.76.
    Explain how a buffer made from ethanoic acid and sodium ethanoate resists a change in pH when a small amount of hydrochloric acid is added.2 marks
  2. Human blood is kept at a pH of about 7.4. This is maintained largely by a buffer system based on carbonic acid, H₂CO₃, and hydrogencarbonate ions, HCO₃⁻.
    Explain why it is important that the pH of blood remains almost constant, and how the buffer achieves this.2 marks
  3. A technician needs 500 cm³ of a buffer solution of pH 5.00. The final buffer must contain ethanoic acid at a concentration of 0.500 mol dm⁻³, and the conjugate base is supplied by solid sodium ethanoate, CH₃COONa (M = 82.0 g mol⁻¹). The Ka of ethanoic acid is 1.74 × 10⁻⁵ mol dm⁻³.
    Calculate the mass of sodium ethanoate needed.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).