Buffer Solutions

What Is a Buffer Solution?

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it. Buffers do not prevent pH changes entirely — they minimise them.

Buffer solutions are essential in biology (blood pH must stay between 7.35 and 7.45), in industrial processes, and throughout analytical chemistry. At A-Level, you need to understand how they work and how to calculate their pH.

Types of Buffer

There are two types:

Acidic buffer (pH < 7): Made from a weak acid and its conjugate base (usually as a salt). For example:

Ethanoic acid (CH₃COOH) + sodium ethanoate (CH₃COONa)
Propanoic acid (CH₃CH₂COOH) + sodium propanoate (CH₃CH₂COONa)
The solution contains both the undissociated weak acid HA and the conjugate base A⁻

Basic buffer (pH > 7): Made from a weak base and its conjugate acid (usually as a salt). For example:

Ammonia (NH₃) + ammonium chloride (NH₄Cl)

Most Edexcel exam questions focus on acidic buffers.

Key principle: A buffer ALWAYS requires a weak acid/base pair. Two strong species cannot form a buffer because a strong acid fully dissociates and cannot provide the HA reservoir needed.

How an Acidic Buffer Works

An acidic buffer contains a large reservoir of:

Weak acid molecules, HA — ready to donate protons
Conjugate base ions, A⁻ — ready to accept protons

The equilibrium in solution is:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

Because both HA and A⁻ are present in large quantities (they are the "reservoir"), the buffer can respond to disturbances.

When acid (H⁺) is added: The extra H⁺ ions react with the conjugate base:

H⁺ + A⁻ → HA

This removes the added H⁺ and converts A⁻ into HA. The pH barely changes because the H⁺ is consumed.

When base (OH⁻) is added: The OH⁻ reacts with the weak acid:

OH⁻ + HA → A⁻ + H₂O

This removes the added OH⁻ and converts HA into A⁻. The pH barely changes because the OH⁻ is consumed.

flowchart LR
    subgraph Buffer Reservoir
        HA["HA (weak acid)"]
        AB["A⁻ (conjugate base)"]
    end
    ACID["Added H⁺"] --> AB
    AB -->|"H⁺ + A⁻ → HA"| HA
    BASE["Added OH⁻"] --> HA
    HA -->|"OH⁻ + HA → A⁻ + H₂O"| AB

Calculating the pH of a Buffer: The Henderson-Hasselbalch Equation

Starting from the Ka expression:

Ka = [H⁺][A⁻] / [HA]

Rearranging for [H⁺]:

[H⁺] = Ka × [HA] / [A⁻]

Taking -log₁₀ of both sides gives the Henderson-Hasselbalch equation:

pH = pKa + log₁₀([A⁻] / [HA])

This is the central equation for buffer pH calculations.

Critical insight: Since both components share the same total volume, the ratio of moles equals the ratio of concentrations. You can use either moles or concentrations in the Henderson-Hasselbalch equation — the volume cancels out.

Worked Example 1: pH of a Buffer from Known Concentrations

Calculate the pH of a buffer made by mixing 100 cm³ of 0.40 mol dm⁻³ ethanoic acid with 100 cm³ of 0.20 mol dm⁻³ sodium ethanoate. Ka for ethanoic acid = 1.74 × 10⁻⁵ mol dm⁻³.

Step 1: Calculate moles.

Moles of CH₃COOH = 0.40 × 0.100 = 0.040 mol Moles of CH₃COO⁻ = 0.20 × 0.100 = 0.020 mol

Step 2: Apply Henderson-Hasselbalch. Since both components are in the same total volume, the ratio of moles equals the ratio of concentrations:

pH = pKa + log₁₀([A⁻] / [HA]) pH = -log₁₀(1.74 × 10⁻⁵) + log₁₀(0.020 / 0.040) pH = 4.76 + log₁₀(0.50) pH = 4.76 + (-0.30) pH = 4.46

Worked Example 2: Using the Ka Method Directly

Some students prefer to rearrange Ka directly rather than use Henderson-Hasselbalch. This is equally valid.

Using the same example:

[H⁺] = Ka × [HA] / [A⁻] [H⁺] = (1.74 × 10⁻⁵) × (0.040 / 0.020) [H⁺] = (1.74 × 10⁻⁵) × 2.0 [H⁺] = 3.48 × 10⁻⁵ mol dm⁻³

pH = -log₁₀(3.48 × 10⁻⁵) = 4.46

Both methods give the same answer. Use whichever you find more natural.

Worked Example 3: Buffer pH with Different Volumes

200 cm³ of 0.30 mol dm⁻³ methanoic acid is mixed with 300 cm³ of 0.10 mol dm⁻³ sodium methanoate. Ka = 1.77 × 10⁻⁴. Calculate the pH.

Moles HCOOH = 0.30 × 0.200 = 0.060 mol Moles HCOO⁻ = 0.10 × 0.300 = 0.030 mol

pH = pKa + log₁₀(0.030 / 0.060) pH = -log₁₀(1.77 × 10⁻⁴) + log₁₀(0.50) pH = 3.75 + (-0.30) pH = 3.45

Note: the total volume is 500 cm³, but since it cancels in the ratio, we do not need it.

Preparing a Buffer with a Specific pH

If you need a buffer at pH 5.00 using ethanoic acid (pKa = 4.76), you can calculate the required ratio:

pH = pKa + log₁₀([A⁻] / [HA]) 5.00 = 4.76 + log₁₀([A⁻] / [HA]) log₁₀([A⁻] / [HA]) = 0.24 [A⁻] / [HA] = 10⁰·²⁴ = 1.74

So you need the conjugate base concentration to be 1.74 times the acid concentration. For example, 0.174 mol dm⁻³ sodium ethanoate with 0.100 mol dm⁻³ ethanoic acid.

Choosing the right weak acid for a target pH: A buffer works best when pH ≈ pKa (effective range is pKa ± 1). To make a buffer at a target pH, choose a weak acid whose pKa is close to that pH.

Target pH	Suitable acid	pKa
3.5 – 4.0	Methanoic acid	3.75
4.5 – 5.0	Ethanoic acid	4.76
4.5 – 5.5	Propanoic acid	4.87
6.0 – 7.0	Carbonic acid (H₂CO₃)	6.37

An Alternative Way to Make a Buffer

You can also make an acidic buffer by partially neutralising a weak acid with a strong base:

CH₃COOH + NaOH → CH₃COONa + H₂O

If you add less NaOH than the stoichiometric amount of CH₃COOH, the resulting solution contains both unreacted CH₃COOH (the weak acid) and the CH₃COO⁻ produced by the reaction (the conjugate base). This is a buffer.

Example: 0.10 mol of CH₃COOH is mixed with 0.040 mol of NaOH. The NaOH reacts with 0.040 mol of CH₃COOH, leaving 0.060 mol of CH₃COOH and producing 0.040 mol of CH₃COO⁻.

pH = 4.76 + log₁₀(0.040 / 0.060) = 4.76 + log₁₀(0.667) = 4.76 - 0.18 = 4.58

Another way buffers form naturally: During a titration of a weak acid with a strong base, the solution acts as a buffer throughout the region before the equivalence point. This is why the titration curve is flat in that region.

Key Points for Exams

A buffer requires both a weak acid AND its conjugate base (or a weak base and its conjugate acid)
Two strong species cannot form a buffer
The Henderson-Hasselbalch equation and the Ka rearrangement are interchangeable
When [A⁻] = [HA], the pH equals the pKa (this is the half-equivalence point on a titration curve)
Buffer pH depends on the ratio [A⁻]/[HA], not on absolute concentrations (though capacity depends on absolute amounts)
You can use moles instead of concentrations in the Henderson-Hasselbalch equation because the volume cancels

Common Exam Mistakes with Buffers

Mistake	Correction
Saying HCl + NaCl is a buffer	HCl is a strong acid — it has no HA reservoir
Forgetting to subtract reacted acid when making a buffer from partial neutralisation	Calculate: remaining HA = initial HA - moles NaOH reacted
Using the total volume in Henderson-Hasselbalch	Volume cancels — you can use the mole ratio directly
Saying buffer pH depends on dilution	Buffer pH depends on the ratio [A⁻]/[HA], not on absolute concentration
Confusing buffer capacity with buffer pH	Capacity depends on absolute moles; pH depends on the ratio
Thinking buffers keep pH constant	Buffers minimise pH changes but do not prevent them

A-Level Deep Dive: Buffer Solutions

Spec mapping

Edexcel 9CH0 specification, Topic 12: Acid-base equilibria, sub-strand 12.4 requires you to define a buffer solution; identify the two compositional types — weak acid + its conjugate base (acidic buffer) and weak base + its conjugate acid (basic buffer); calculate buffer pH using Ka and the Henderson-Hasselbalch equation pH = pKa + log₁₀([A⁻]/[HA]); recognise physiological and pharmaceutical buffer applications including the carbonic-acid / hydrogen-carbonate system in blood (refer to the official specification document for exact wording). Examined in Paper 1 (Topic 12 conceptual + buffer pH calculations) and Paper 3 (synoptic, including biological/pharmaceutical contexts).

Worked example with full mark scheme

Question (8 marks):

(a) State what is meant by a buffer solution and describe the two compositional types with one example of each. (3)

(b) A buffer is prepared by mixing 50.0 cm³ of 0.20 mol dm⁻³ ethanoic acid (pKa = 4.76) with 25.0 cm³ of 0.20 mol dm⁻³ sodium ethanoate. Calculate the pH of the buffer at 298 K. (3)

(c) Calculate the ratio [A⁻]/[HA] needed to prepare an ethanoate buffer of pH 5.20 from the same starting materials. (2)

Solution with mark scheme:

(a) A buffer is a solution that resists changes in pH when small amounts of acid or base are added (or when it is diluted within reasonable limits).

Type 1: Acidic buffer — a weak acid + its conjugate base, e.g. CH₃COOH / CH₃COONa.

Type 2: Basic buffer — a weak base + its conjugate acid, e.g. NH₃ / NH₄Cl.

M1 — definition includes "resists pH change" + "small amounts of acid/base". M1 — names both compositional types. A1 — gives one valid example for each.

(b) Moles ethanoic acid (HA) = 0.050 × 0.20 = 1.00 × 10⁻² mol.

Moles ethanoate (A⁻) = 0.025 × 0.20 = 5.00 × 10⁻³ mol.

Total volume = 0.075 dm³ (cancels in the ratio).

Ratio [A⁻]/[HA] = 5.00 × 10⁻³ / 1.00 × 10⁻² = 0.50.

pH = pKa + log₁₀([A⁻]/[HA]) = 4.76 + log₁₀(0.50) = 4.76 − 0.30 = 4.46.

M1 — moles of HA and A⁻. M1 — Henderson-Hasselbalch substitution. A1 — pH = 4.46.

[A⁻]/[HA] = 10⁰·⁴⁴ = 2.75.

M1 — correct rearrangement. A1 — ratio = 2.75 (or 2.8 to 2 s.f.).

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks): Identify which of the following solutions act as buffers and explain your reasoning in each case.

Mixture	Composition
A	50 cm³ of 0.10 mol dm⁻³ HCl + 50 cm³ of 0.10 mol dm⁻³ NaCl
B	50 cm³ of 0.10 mol dm⁻³ CH₃COOH + 25 cm³ of 0.10 mol dm⁻³ NaOH
C	50 cm³ of 0.10 mol dm⁻³ NH₃ + 50 cm³ of 0.10 mol dm⁻³ NH₄Cl
D	50 cm³ of 0.10 mol dm⁻³ HCl + 50 cm³ of 0.10 mol dm⁻³ NaOH

Mark scheme decomposition by AO:

B1 (AO2.1) — A: not a buffer; HCl is a strong acid, Cl⁻ is not its conjugate base in any acid-base sense (Cl⁻ has no buffering capacity).
B1 (AO2.1) — B: is a buffer; partial neutralisation of weak acid by strong base produces a mixture of CH₃COOH (excess) and CH₃COO⁻ (the salt formed). Half-neutralisation gives equal moles, but here only one quarter is neutralised, so [HA] > [A⁻].
B1 (AO2.1) — C: is a buffer; weak base NH₃ + its conjugate acid NH₄⁺ (from NH₄Cl). Classic basic-buffer composition.
B1 (AO2.1) — D: not a buffer; stoichiometric strong acid + strong base gives only NaCl(aq) at pH 7, with no weak conjugate pair to resist further pH change.
B1 (AO1.2) — generalisation: a buffer requires a weak species and its conjugate; both must be present in significant concentrations.
B1 (AO2.5) — closing comment: HCl + NaCl is the classic distractor — students often think any acid + salt is a buffer, but the salt must contain the conjugate base of the weak acid.

Total: 6 marks all AO2/AO3. Edexcel uses identification questions to test whether candidates have internalised the conjugate-pair requirement.

Synoptic links

Connects to:

Topic 12.3 — Ka (Lesson 3): the Henderson-Hasselbalch equation is a direct rearrangement of Ka = [H⁺][A⁻]/[HA] taking logs of both sides. Confidence with Ka here is essential.
Topic 12.5 — Titration curves (Lesson 8): the buffering region of a weak acid / strong base titration (between ≈ 10% and 90% titrant) is exactly where the Henderson-Hasselbalch equation applies. Half-equivalence is the mid-buffer point at pH = pKa.
Biology (blood pH homeostasis): the H₂CO₃ / HCO₃⁻ buffer system maintains arterial blood pH at 7.35–7.45 against metabolic acid loading. CO₂ exhalation regulates [H₂CO₃]; renal HCO₃⁻ resorption regulates [HCO₃⁻].