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.
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.
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.
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.
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.
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
- 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
- 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
- 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
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).