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A buffer solution is a mixture that resists changes in pH when small amounts of strong acid or strong base are added, or when the mixture is diluted modestly. Buffers are not a single substance; they are deliberate combinations of a weak acid with its conjugate base, or a weak base with its conjugate acid, both present in significant and comparable concentrations. The chemistry of buffers sits at the intersection of equilibrium (§3.1.6), acid-base theory (lesson 1 of this course), and the quantitative pH treatment developed for weak acids (lesson 3). In this lesson we define the two buffer types, derive the Henderson-Hasselbalch equation from the Ka expression, work through buffer pH calculations both at preparation and after small additions of strong acid, examine buffer capacity and the ±1 pKa rule, and explore biological buffers — the carbonate system in blood plasma, the phosphate buffer that dominates inside cells, and the role of protein side chains in maintaining physiological pH. Required Practical 9 (lesson 7) returns to the experimental side.
Spec mapping (AQA 7405): This lesson maps to §3.1.12 (acids and bases), specifically the buffer-action and Henderson-Hasselbalch content in the second half of that section. It builds directly on lesson 3 of this course (pH of weak acids — the Ka expression and the standard weak-acid approximation), connects to lesson 5 (pH curves and indicators — the buffer region of a weak-acid/strong-base curve is the same physics treated here), and previews lesson 7 (Required Practical 9 — half-equivalence-point determination of pKa). Equilibrium (§3.1.6) underpins the entire treatment: a buffer works precisely because Le Chatelier's principle dictates how the weak-acid dissociation responds to added H⁺ or OH⁻. Refer to the official AQA specification document for the exact wording of each subsection.
Assessment objectives: Defining a buffer and stating the Henderson-Hasselbalch equation are AO1 recall items. Computing the pH of a buffer at a given component ratio, and the new pH after a small addition of strong acid or strong base, are core AO2 calculations that appear on most Paper 2 papers. Evaluating buffer capacity, choosing a weak acid with appropriate pKa for a target pH window, and explaining the biological significance of the carbonate buffer system involve AO3 reasoning and synoptic synthesis with respiration physiology.
Key Definition: A buffer solution is a solution that resists changes in pH when small quantities of strong acid or strong base are added, or when the solution is diluted modestly. The word "resists" is deliberate: a buffer does not prevent pH change entirely; it limits the change to a small fraction of what would occur in pure water or in an unbuffered electrolyte.
Acidic buffer (typically pH 3–7): a weak acid HA together with its conjugate base A⁻, supplied as a salt of the acid. The canonical A-Level example is ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COONa), which together buffer near pH 4.76 (the pKa of ethanoic acid). Other examples include the methanoic acid / sodium methanoate pair (pKa 3.75) and the benzoic acid / sodium benzoate pair (pKa 4.20).
Basic buffer (typically pH 8–11): a weak base B together with its conjugate acid BH⁺, supplied as a salt of the base. The canonical example is ammonia (NH₃) and ammonium chloride (NH₄Cl), which together buffer near pH 9.25 (since pKb of NH₃ ≈ 4.75 and pKa of NH₄⁺ ≈ 9.25 — and 14 − 4.75 = 9.25, as Ka·Kb = Kw demands). The trimethylamine / trimethylammonium chloride pair gives buffering near pH 9.8.
A buffer is therefore a deliberate mixture, not a single compound. Pure ethanoic acid alone is not a buffer; pure sodium ethanoate alone is not a buffer; the combination at significant concentrations of both is.
The buffer contains a large reservoir of both HA (the weak acid, acting as the proton donor and as the species that mops up added OH⁻) and A⁻ (the conjugate base, acting as the species that mops up added H⁺). Both reservoirs must be present in significant and comparable concentrations for buffering to operate.
When a small amount of strong acid (H⁺) is added:
The added H⁺ ions are rapidly consumed by the conjugate base:
A⁻ + H⁺ → HA
For an ethanoic-acid buffer: CH₃COO⁻ + H⁺ → CH₃COOH. The added protons are converted into more weak acid; they do not contribute directly to the [H⁺] of the solution because A⁻ has a much higher affinity for H⁺ than water does (this is precisely what "weak acid" means at equilibrium). The [HA]/[A⁻] ratio shifts slightly in favour of HA, but because both quantities are large, the logarithm of the ratio shifts only slightly — and the Henderson-Hasselbalch equation shows pH depends on the log of the ratio, not on its absolute value.
When a small amount of strong base (OH⁻) is added:
The added OH⁻ ions are rapidly consumed by the weak acid:
HA + OH⁻ → A⁻ + H₂O
For an ethanoic-acid buffer: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The hydroxide ions are converted into water and additional conjugate base; they do not survive to raise [OH⁻] (and hence to lower [H⁺] via Kw) appreciably. The [HA]/[A⁻] ratio shifts slightly in favour of A⁻, but again the log-ratio change — and hence the pH change — is small.
The equilibrium view. From Le Chatelier's principle applied to HA ⇌ H⁺ + A⁻: adding H⁺ shifts the equilibrium to the left (forming HA); adding OH⁻ removes H⁺ from the equilibrium (by reacting OH⁻ + H⁺ → H₂O), shifting it to the right (forming more A⁻). In both cases the system absorbs the perturbation and the change in [H⁺] is much smaller than would result from adding the same H⁺ or OH⁻ to pure water of the same volume. The reservoirs of HA and A⁻ act as a chemical shock absorber.
Starting from the Ka expression for the weak acid:
Ka = [H⁺][A⁻] / [HA]
Rearranging to isolate [H⁺]:
[H⁺] = Ka × [HA] / [A⁻]
Taking −log₁₀ of both sides gives the Henderson-Hasselbalch equation (named after the American biochemist Henderson and the Danish physician Hasselbalch, who developed the logarithmic form in the early twentieth century):
pH = pKa + log₁₀([A⁻] / [HA])
For a basic buffer, the corresponding form is:
pOH = pKb + log₁₀([BH⁺] / [B]), with pH = 14.00 − pOH at 298 K.
Three features of the equation deserve attention:
A buffer is made by dissolving 0.100 mol of sodium ethanoate (CH₃COONa) in 1.00 dm³ of 0.100 mol dm⁻³ ethanoic acid. Ka(CH₃COOH) = 1.74 × 10⁻⁵ mol dm⁻³. Calculate the pH.
[HA] = [CH₃COOH] = 0.100 mol dm⁻³ [A⁻] = [CH₃COO⁻] = 0.100 mol dm⁻³ Ratio [A⁻]/[HA] = 1.00, so log₁₀(1) = 0.
pKa = −log₁₀(1.74 × 10⁻⁵) = 4.76
pH = pKa + log₁₀(1) = 4.76
A 1:1 acidic buffer always sits at pH = pKa. This is the half-equivalence point of the corresponding weak-acid / strong-base titration (lesson 5 of this course) and is the experimental basis for determining pKa from a pH curve (lesson 7).
A buffer is made from 0.40 mol dm⁻³ ethanoic acid and 0.20 mol dm⁻³ sodium ethanoate. Calculate the pH using Ka = 1.74 × 10⁻⁵.
Ratio [A⁻]/[HA] = 0.20 / 0.40 = 0.500 log₁₀(0.500) = −0.301
pH = 4.76 + (−0.301) = 4.46
Alternatively, working from [H⁺] directly:
[H⁺] = Ka × [HA]/[A⁻] = 1.74 × 10⁻⁵ × (0.40/0.20) = 3.48 × 10⁻⁵ pH = −log₁₀(3.48 × 10⁻⁵) = 4.46
Both routes give the same answer (as they must — they are algebraically equivalent). Doubling the acid (or halving the salt) drops the pH by exactly log₁₀(2) = 0.301 units.
100 cm³ of 0.100 mol dm⁻³ CH₃COOH is mixed with 100 cm³ of 0.100 mol dm⁻³ CH₃COONa. Step 1: calculate the initial pH. Step 2: calculate the pH after adding 5.00 cm³ of 1.00 mol dm⁻³ HCl. Ka = 1.74 × 10⁻⁵.
Step 1 — initial buffer:
Moles CH₃COOH = 0.100 × 0.100 = 0.0100 mol Moles CH₃COO⁻ = 0.100 × 0.100 = 0.0100 mol Ratio = 1:1 → pH = pKa = 4.76.
Step 2 — after adding 5.00 cm³ of 1.00 mol dm⁻³ HCl:
Moles H⁺ added = 0.00500 × 1.00 = 5.00 × 10⁻³ mol = 0.00500 mol.
The added H⁺ reacts with CH₃COO⁻: CH₃COO⁻ + H⁺ → CH₃COOH. So:
Moles CH₃COOH after = 0.0100 + 0.00500 = 0.0150 mol Moles CH₃COO⁻ after = 0.0100 − 0.00500 = 0.00500 mol
The new total volume is 200 + 5.00 = 205 cm³, but because pH depends on the ratio rather than the absolute concentrations, we can use moles directly:
Ratio [A⁻]/[HA] = 0.00500 / 0.0150 = 0.333 log₁₀(0.333) = −0.477
pH = 4.76 + (−0.477) = 4.28
The pH has changed by 4.76 − 4.28 = 0.48 units. For comparison, adding the same 5.00 × 10⁻³ mol of HCl to 200 cm³ of pure water (initial pH 7.00) would give [H⁺] ≈ 5.00 × 10⁻³ / 0.205 = 0.0244 mol dm⁻³, pH = 1.61 — a change of more than 5 pH units. The buffer absorbs the addition with roughly a tenth of the pH swing.
50.0 cm³ of 0.100 mol dm⁻³ CH₃COOH is partially neutralised by 30.0 cm³ of 0.100 mol dm⁻³ NaOH. Calculate the pH of the resulting mixture. Ka = 1.74 × 10⁻⁵.
Moles CH₃COOH initial = 0.0500 × 0.100 = 5.00 × 10⁻³ mol Moles NaOH = 0.0300 × 0.100 = 3.00 × 10⁻³ mol
NaOH neutralises CH₃COOH: CH₃COOH + NaOH → CH₃COONa + H₂O
After reaction: Moles CH₃COOH = 5.00 × 10⁻³ − 3.00 × 10⁻³ = 2.00 × 10⁻³ mol Moles CH₃COO⁻ = 3.00 × 10⁻³ mol
Ratio [A⁻]/[HA] = 3.00 / 2.00 = 1.50; log₁₀(1.50) = 0.176
pH = 4.76 + 0.176 = 4.94
The mixture is a buffer because both CH₃COOH and CH₃COO⁻ are present in significant amounts. This is the buffer region of a weak-acid / strong-base titration (lesson 5).
The phosphate buffer at physiological pH uses the H₂PO₄⁻ / HPO₄²⁻ conjugate pair. The relevant pKa is pKa2 = 7.20 (the second dissociation of phosphoric acid, H₂PO₄⁻ ⇌ HPO₄²⁻ + H⁺). What ratio of [HPO₄²⁻] / [H₂PO₄⁻] gives a buffer of pH 7.40 (the pH of blood plasma)?
7.40 = 7.20 + log₁₀([HPO₄²⁻]/[H₂PO₄⁻]) log₁₀(ratio) = 0.20 ratio = 10^0.20 = 1.58
So roughly 1.6 mol HPO₄²⁻ per mol H₂PO₄⁻ achieves pH 7.40. This is the dominant intracellular buffer; in red blood cells and most cells the phosphate pair maintains cytoplasmic pH near 7.2–7.4.
A buffer's effectiveness is quantified by its buffer capacity (β), defined informally as the moles of strong acid or strong base required to change the pH of 1 dm³ of buffer by one pH unit. Two factors determine capacity:
The practical ±1 pKa rule follows: a buffer is effective in the range pH = pKa ± 1. Outside this window, one reservoir is at most one-tenth the size of the other, and added acid or base of significant amount will exhaust the smaller reservoir and break the buffer. For ethanoic acid (pKa 4.76), the useful range is roughly pH 3.76–5.76; for the dihydrogenphosphate / hydrogenphosphate system (pKa2 = 7.20), the range is roughly pH 6.20–8.20, covering all physiological values.
The design rule is straightforward: pick a weak acid whose pKa is closest to the target pH, then adjust the [A⁻]/[HA] ratio (via the Henderson-Hasselbalch equation) to fine-tune. A short table of common laboratory buffer systems:
| Conjugate pair | pKa | Useful pH range |
|---|---|---|
| HCOOH / HCOO⁻ (methanoic) | 3.75 | 2.8–4.8 |
| CH₃COOH / CH₃COO⁻ (ethanoic) | 4.76 | 3.8–5.8 |
| H₂CO₃ / HCO₃⁻ (carbonic, pKa1) | 6.35 | 5.4–7.4 |
| H₂PO₄⁻ / HPO₄²⁻ (phosphate, pKa2) | 7.20 | 6.2–8.2 |
| NH₄⁺ / NH₃ (ammonium) | 9.25 | 8.3–10.3 |
| HCO₃⁻ / CO₃²⁻ (carbonate, pKa2) | 10.33 | 9.3–11.3 |
To prepare a buffer at pH 5.00, ethanoic acid (pKa 4.76) is the natural choice; using the Henderson-Hasselbalch equation, [A⁻]/[HA] = 10^(5.00 − 4.76) = 10^0.24 = 1.74. Mixing 1.74 mol of sodium ethanoate with 1.00 mol of ethanoic acid in 1 dm³ gives pH 5.00.
Mammalian blood plasma is buffered at pH 7.35–7.45. The principal extracellular buffer is the carbonic acid / hydrogencarbonate (bicarbonate) system:
CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺
At first sight this system looks misallocated: the relevant pKa1 of carbonic acid is approximately 6.35, well below physiological pH 7.4, so the ±1 pKa rule predicts that the carbonate buffer should be ineffective at pH 7.4 (the [HCO₃⁻]/[H₂CO₃] ratio is about 10^1.05 ≈ 11:1, far from 1:1). It works in vivo for two reasons:
The [HCO₃⁻] reservoir is large — about 24 mmol dm⁻³ in plasma, much higher than dissolved CO₂ (~1.2 mmol dm⁻³). The absolute reservoir size compensates for the unfavourable ratio.
The system is open, not closed. Dissolved CO₂ is in equilibrium with gaseous CO₂ in the alveoli of the lungs, and respiratory rate can adjust CO₂ partial pressure in seconds-to-minutes. If blood pH drops (acidosis), respiratory rate increases, exhaling CO₂ and shifting the equilibrium left; if blood pH rises (alkalosis), respiratory rate decreases, retaining CO₂ and shifting right. The kidneys provide a slower (hours-to-days) adjustment of [HCO₃⁻] by reabsorption or excretion. Together, the lungs and kidneys make the carbonate buffer dynamically open — capacity is effectively unlimited on physiological timescales.
The condition where blood pH falls below 7.35 is called acidosis; above 7.45 is alkalosis. Either represents a life-threatening failure of the carbonate buffer system and the lung-kidney regulation that supports it.
The phosphate buffer (H₂PO₄⁻ / HPO₄²⁻, pKa2 = 7.20) dominates intracellular pH regulation, where phosphate concentrations are higher than in plasma. Its pKa sits almost exactly at physiological pH, so capacity is maximised. Protein buffers — especially the imidazole side chain of histidine (pKa ≈ 6.0) and the amino and carboxyl groups of polypeptide chains — provide additional buffering, particularly important in haemoglobin, which carries protons released by CO₂ hydration in tissues to be exhaled in the lungs (the Bohr effect). Together, the carbonate, phosphate, and protein buffer systems give blood plasma a remarkable resistance to pH disturbance.
To make 250 cm³ of a pH 4.76 ethanoic-acid buffer at 0.100 mol dm⁻³ total concentration (i.e. 0.0500 mol dm⁻³ CH₃COOH + 0.0500 mol dm⁻³ CH₃COONa):
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