Buffer Solutions: How They Work

Spec Mapping — OCR H432 Module 5.1.3 — Acids, bases and buffers, sub-topic on buffer solutions: the definition of a buffer, composition of an acidic buffer (weak acid + conjugate base) and basic buffer (weak base + conjugate acid), preparation by direct mixing or partial neutralisation, mechanism of buffer action via Le Chatelier on the dissociation equilibrium, response to added H⁺ and OH⁻, buffer capacity and useful pH range, and important biological / industrial examples especially the HCO₃⁻ / H₂CO₃ system that buffers blood at pH 7.4 (refer to the official OCR H432 specification document for exact wording). This is the conceptual lesson; the quantitative Henderson-Hasselbalch calculation is in lesson 8.

A buffer solution is one of the most physiologically and industrially important pieces of chemistry in the syllabus — blood, cell cytoplasm, fermentation broths, shampoos, soft drinks and almost every quantitative biochemical assay rely on careful pH buffering to maintain activity. The mechanism is the central concept: a weak acid HA and its conjugate base A⁻ coexist in equilibrium, providing reservoirs that absorb added H⁺ (via A⁻ + H⁺ → HA) or added OH⁻ (via HA + OH⁻ → A⁻ + H₂O) without large changes in [H⁺] because the ratio $[\text{A}^-]/[\text{HA}]$[A−]/[HA] — the key determinant of pH — barely shifts. This lesson establishes the qualitative understanding; lesson 8 derives Henderson-Hasselbalch and uses it for quantitative pH-change calculations.

Key Equilibrium: $\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{A}^-\text{(aq)}$HA(aq)⇌H+(aq)+A−(aq) $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \quad \Rightarrow \quad [\text{H}^+] = K_a \cdot \frac{[\text{HA}]}{[\text{A}^-]}$Ka​=[HA][H+][A−]​⇒[H+]=Ka​⋅[A−][HA]​ The buffer pH is set by the ratio [HA] / [A⁻]; when small amounts of H⁺ or OH⁻ are added, both [HA] and [A⁻] shift by small fractional amounts, so the ratio (and pH) barely changes.

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What is a buffer?

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it, or when it is diluted. The keyword is resists — buffers do not prevent pH changes, they minimise them. When a buffer is overwhelmed by large quantities of H⁺ or OH⁻, its capacity is exceeded and the pH then changes sharply (as on either side of the central horizontal region of a weak-acid / strong-base titration curve).

Buffer-controlled pH is essential wherever a biological or chemical process must occur within a narrow pH window:

• Human blood is buffered at pH 7.35–7.45 by the HCO₃⁻ / H₂CO₃ system, with secondary contributions from HPO₄²⁻ / H₂PO₄⁻ and from buffering proteins such as haemoglobin.
• Cell cytoplasm is buffered around pH 7.0–7.2 — most cellular enzymes operate within a 0.5 pH-unit window.
• Saliva is buffered around pH 6.5–7.5 by HCO₃⁻ and phosphate; saliva flow rate spikes after meals partly to dilute and re-buffer acidic food.
• Stomach acid is buffered at pH ~1.5 by HCl plus dissolved mucin glycoproteins.
• Shampoos and conditioners are buffered to match skin pH (~5.5) to prevent irritation.
• Industrial fermentations use phosphate or citrate buffers to optimise enzyme activity.
• Analytical chemistry depends on buffered calibration solutions for pH-meter calibration (pH 4.00, 7.00, 10.00) and on buffered eluents in HPLC and ion chromatography.

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Composition of an acidic buffer

An acidic buffer (pH < 7) consists of two components:

1. A weak acid HA — to react with added OH⁻ (via HA + OH⁻ → A⁻ + H₂O).
2. The conjugate base A⁻ — usually supplied as the sodium or potassium salt of the acid, to react with added H⁺ (via A⁻ + H⁺ → HA).

Typical examples and their buffering ranges:

Weak acid	$\text{p}K_a$pKa​	Salt providing A⁻	Useful buffer pH range
HCOOH	3.80	HCOONa (sodium methanoate)	2.8 – 4.8
CH₃COOH	4.76	CH₃COONa (sodium ethanoate)	3.8 – 5.8
C₆H₅COOH	4.20	C₆H₅COONa (sodium benzoate)	3.2 – 5.2
H₂CO₃	6.37	NaHCO₃ (sodium hydrogencarbonate)	5.4 – 7.4
H₂PO₄⁻	7.21	HPO₄²⁻ (Na₂HPO₄)	6.2 – 8.2 (KEY for blood)
HCO₃⁻	10.33	CO₃²⁻ (Na₂CO₃)	9.3 – 11.3

A basic buffer (pH > 7) consists of a weak base and its conjugate acid — e.g. NH₃ / NH₄Cl ($\text{p}K_a$pKa​ of NH₄⁺ = 9.25). OCR Module 5.1.3 focuses primarily on acidic buffers but expects awareness of the basic-buffer analogue.

Preparing an acidic buffer

Two practical preparation methods:

Method 1 — direct mixing of weak acid + salt

Dissolve a measured amount of the weak acid and its sodium salt together. For a 1:1 ethanoate buffer: mix 0.100 mol of CH₃COOH and 0.100 mol of CH₃COONa in 1 dm³ of water. The resulting solution has $[\text{HA}] = [\text{A}^-] = 0.100$[HA]=[A−]=0.100 mol dm⁻³ and pH = $\text{p}K_a$pKa​ = 4.76 (a 1:1 buffer is centred at $\text{p}K_a$pKa​).

Method 2 — partial neutralisation of a weak acid with a strong base

Start with a known amount of the weak acid and add less than the stoichiometric amount of strong base (NaOH). The base neutralises some of the HA to A⁻, leaving the remaining HA un-reacted:

$\text{HA} + \text{NaOH} \rightarrow \text{Na}^+ + \text{A}^- + \text{H}_2\text{O}$HA+NaOH→Na++A−+H2​O

For example, adding 0.50 mol NaOH to 1.00 mol CH₃COOH produces a buffer containing 0.50 mol of CH₃COOH (un-reacted) and 0.50 mol of CH₃COO⁻ (newly produced) — a 1:1 buffer at pH = $\text{p}K_a$pKa​ = 4.76.

The elegant special case is half-neutralisation: when exactly half the weak acid has been converted to its salt, $[\text{HA}] = [\text{A}^-]$[HA]=[A−] and pH = $\text{p}K_a$pKa​. This is the basis of the half-equivalence-point analysis in lesson 9, and is the principal experimental method for measuring $\text{p}K_a$pKa​ of an unknown weak acid.

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The buffering equilibrium — Le Chatelier in action

Consider an ethanoic acid / ethanoate buffer. Both components coexist in the same solution and share the equilibrium:

$\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CH}_3\text{COO}^-\text{(aq)}$CH3​COOH(aq)⇌H+(aq)+CH3​COO−(aq)

The weak acid is the reservoir of undissociated HA (ready to neutralise added OH⁻), and the salt provides the reservoir of A⁻ (ready to neutralise added H⁺). The pH is set by the ratio $[\text{HA}]:[\text{A}^-]$[HA]:[A−] and remains nearly constant as small quantities of acid or base are added.

graph TD
  A["Add H+"] -->|"reacts with A-"| B["Forms HA"]
  C["Add OH-"] -->|"reacts with HA"| D["Forms A- + H2O"]
  E["[H+] stays nearly constant"]
  B --> E
  D --> E
  F["Ratio [HA]/[A-] changes only slightly"] --> E

Response to added acid (H⁺)

When a small amount of strong acid is added:

$\text{CH}_3\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{CH}_3\text{COOH(aq)}$CH3​COO−(aq)+H+(aq)→CH3​COOH(aq)

• $[\text{A}^-]$[A−] decreases slightly (some converted to HA).
• $[\text{HA}]$[HA] increases slightly.
• $[\text{H}^+]$[H+] rises only very slightly because the bulk of the added H⁺ is mopped up by the large A⁻ reservoir.

In Le Chatelier terms: adding H⁺ to the system $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$HA⇌H++A− shifts the equilibrium to the left, consuming both the added H⁺ and some A⁻, regenerating HA. The position of the equilibrium adjusts to restore $[\text{H}^+]$[H+] close to its pre-addition value.

Response to added base (OH⁻)

When a small amount of strong base is added:

$\text{CH}_3\text{COOH(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_2\text{O(l)}$CH3​COOH(aq)+OH−(aq)→CH3​COO−(aq)+H2​O(l)

• $[\text{HA}]$[HA] decreases slightly.
• $[\text{A}^-]$[A−] increases slightly.
• $[\text{OH}^-]$[OH−] is consumed; $[\text{H}^+]$[H+] is only very slightly reduced because HA dissociates further to replace lost H⁺.

In Le Chatelier terms: OH⁻ removes H⁺ from the system, shifting the equilibrium to the right. HA dissociates to replace some of the lost H⁺, partially restoring it.

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Why it works — the ratio argument

The key quantitative insight (the basis of Henderson-Hasselbalch in lesson 8): for a buffer at equilibrium,

$[\text{H}^+] = K_a \cdot \frac{[\text{HA}]}{[\text{A}^-]}$[H+]=Ka​⋅[A−][HA]​

so the pH depends on the ratio [HA]/[A⁻] (logarithmically). Because both [HA] and [A⁻] are present in large and comparable quantities (typically 0.05 to 0.5 mol dm⁻³), adding a small amount (say 0.001 mol) of H⁺ or OH⁻ shifts each component by only a few percent — and the ratio by even less. Since the ratio barely changes, neither does the pH.

Contrast with unbuffered water: adding 0.001 mol of HCl to 1 dm³ of pure water changes $[\text{H}^+]$[H+] from $10^{-7}$10−7 to $10^{-3}$10−3 — a pH drop of 4.00 units. Adding the same amount to a 0.10 mol dm⁻³ ethanoate buffer shifts pH by only ~0.02 units. Three orders of magnitude better resistance to acid.

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Buffer capacity and useful pH range

Buffer capacity $\beta$β is the moles of acid or base a buffer can absorb per pH unit change. It depends on two factors:

1. Total concentration of buffer components. A 1.0 mol dm⁻³ buffer has 10× the capacity of a 0.1 mol dm⁻³ buffer at the same ratio.
2. Ratio of [HA] to [A⁻]. Capacity is maximal when [HA] = [A⁻] (pH = $\text{p}K_a$pKa​) and falls off as the ratio moves away from 1:1. At [HA]:[A⁻] = 10:1 (pH = $\text{p}K_a$pKa​ − 1) or 1:10 (pH = $\text{p}K_a$pKa​ + 1), capacity is roughly halved.

Useful pH range: a good rule of thumb is that a buffer is effective within ±1 pH unit of its $\text{p}K_a$pKa​. Outside this range one component dominates and the buffer capacity is poor. For ethanoic acid ($\text{p}K_a = 4.76$pKa​=4.76), the ethanoate buffer works well from pH 3.76 to 5.76. To buffer at pH 7, you would choose a weak acid with $\text{p}K_a$pKa​ near 7 — H₂PO₄⁻ ($\text{p}K_{a2} = 7.21$pKa2​=7.21) is the natural choice, which is also why phosphate buffering is so important in cell biology.

When a buffer is asked to absorb more acid or base than it can handle, the component being attacked is depleted, the ratio shifts dramatically, and the pH crashes (acid added) or spikes (base added). This is the breakdown of buffer action — see the steep flanks on either side of the central plateau in a weak-acid / strong-base titration curve (lesson 9). A 0.10 mol dm⁻³ ethanoate buffer in 100 cm³ contains 0.010 mol of each component; it can absorb perhaps 0.005 mol of added H⁺ (half the A⁻) before the pH starts to fall noticeably. Beyond that, the buffer is overwhelmed and behaves like a dilute weak-acid solution.

The choice of buffer for an experiment is therefore a four-way trade-off: target pH (sets $\text{p}K_a$pKa​ within ±1), required capacity (sets total concentration), ionic strength compatibility (some assays are intolerant of high salt), and chemical inertness (the buffer must not react with the experimental system — phosphate precipitates Ca²⁺; carbonate dissolves CO₂ from air; Tris reacts with copper). Biochemistry lectures spend whole sessions on buffer choice.

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Blood buffering — the carbonic acid / hydrogencarbonate system

Human blood is maintained at pH 7.35–7.45 — a range of just 0.1 units. Going outside this range (acidosis below 7.35; alkalosis above 7.45) causes serious physiological problems (enzyme malfunction, oxygen-binding errors in haemoglobin, neuromuscular irritability). The main blood buffer is the carbonic acid / hydrogencarbonate system:

$\text{H}_2\text{CO}_3\text{(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{HCO}_3^-\text{(aq)}, \quad K_a = 4.3 \times 10^{-7} \text{ mol dm}^{-3}, \quad \text{p}K_a = 6.37$H2​CO3​(aq)⇌H+(aq)+HCO3−​(aq),Ka​=4.3×10−7 mol dm−3,pKa​=6.37

At blood pH 7.4, the Henderson-Hasselbalch equation gives $[\text{HCO}_3^-]/[\text{H}_2\text{CO}_3] \approx 11:1$[HCO3−​]/[H2​CO3​]≈11:1. That ratio is far from the ideal 1:1 (which would maximise capacity), so by the "rule of thumb" the blood buffer is operating somewhat outside its peak window. The system survives because:

1. Both concentrations are high. $[\text{HCO}_3^-] \approx 24$[HCO3−​]≈24 mmol dm⁻³ in plasma; $[\text{H}_2\text{CO}_3] \approx 1.2$[H2​CO3​]≈1.2 mmol dm⁻³.
2. The buffer is coupled to respiration. $\text{H}_2\text{CO}_3 \rightleftharpoons \text{CO}_2\text{(aq)} + \text{H}_2\text{O}$H2​CO3​⇌CO2​(aq)+H2​O — the dissolved CO₂ is in rapid exchange with alveolar CO₂. Exhaling CO₂ removes H₂CO₃, shifting the buffer equilibrium left and raising the [HCO₃⁻]/[H₂CO₃] ratio (so pH rises). Retaining CO₂ does the opposite.
3. Secondary buffers. Plasma proteins (especially the histidine residues of haemoglobin) and the phosphate system (HPO₄²⁻ / H₂PO₄⁻, $\text{p}K_a$pKa​ 7.21) provide additional capacity.