How Buffers Resist pH Changes

The Mechanism in Detail

In the previous lesson you learned the equations describing buffer action. Now let us examine the mechanism more deeply, including the role of Le Chatelier's principle, buffer capacity, and real-world applications.

An acidic buffer contains a weak acid HA and its conjugate base A⁻ in equilibrium:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

The key to buffer action is that both HA and A⁻ are present in large, comparable quantities. They act as reservoirs that can absorb added acid or base without significant changes in pH.

Adding Acid to a Buffer: Step-by-Step

When a small amount of strong acid (H⁺) is added to the buffer:

The added H⁺ ions increase [H⁺] momentarily
The conjugate base A⁻ reacts with the added H⁺: H⁺ + A⁻ → HA
This removes the added H⁺ from solution
[HA] increases slightly while [A⁻] decreases slightly
The ratio [HA]/[A⁻] changes only marginally because both are present in large amounts
pH decreases only very slightly

From Le Chatelier's perspective, adding H⁺ to the system disturbs the equilibrium. The system responds by shifting the equilibrium to the left (towards HA), consuming the added H⁺ and restoring the position close to the original.

Exam marking point: When explaining buffer action for added acid, you must state:

The added H⁺ reacts with A⁻ (the conjugate base)
The equation: H⁺ + A⁻ → HA
This removes the added H⁺ so pH changes very little

Adding Base to a Buffer: Step-by-Step

When a small amount of strong base (OH⁻) is added:

The OH⁻ reacts directly with the weak acid HA: OH⁻ + HA → A⁻ + H₂O
This removes the added OH⁻ from solution
[HA] decreases slightly while [A⁻] increases slightly
The ratio changes marginally
pH increases only very slightly

Exam marking point: When explaining buffer action for added base, you must state:

The added OH⁻ reacts with HA (the weak acid)
The equation: OH⁻ + HA → A⁻ + H₂O
This removes the added OH⁻ so pH changes very little

Worked Example: pH Change When Acid Is Added

A buffer contains 0.10 mol of CH₃COOH and 0.10 mol of CH₃COO⁻ in 1.0 dm³. Ka = 1.74 × 10⁻⁵ mol dm⁻³. Calculate the pH before and after adding 0.010 mol of HCl.

Before:

pH = pKa + log₁₀(0.10 / 0.10) = 4.76 + 0 = 4.76

After adding HCl:

The 0.010 mol of H⁺ reacts with 0.010 mol of CH₃COO⁻:

New moles of CH₃COO⁻ = 0.10 - 0.010 = 0.090 mol New moles of CH₃COOH = 0.10 + 0.010 = 0.110 mol

pH = 4.76 + log₁₀(0.090 / 0.110) pH = 4.76 + log₁₀(0.818) pH = 4.76 + (-0.087) pH = 4.67

The pH has changed by only 0.09 units despite adding 0.010 mol of strong acid. Without the buffer, 0.010 mol of HCl in 1.0 dm³ would give pH = 2.00 — a change of 2.76 units from the original.

Worked Example: pH Change When Base Is Added

Using the same buffer (0.10 mol CH₃COOH, 0.10 mol CH₃COO⁻, 1.0 dm³), calculate the pH after adding 0.010 mol NaOH.

The 0.010 mol of OH⁻ reacts with 0.010 mol of CH₃COOH:

New moles of CH₃COOH = 0.10 - 0.010 = 0.090 mol New moles of CH₃COO⁻ = 0.10 + 0.010 = 0.110 mol

pH = 4.76 + log₁₀(0.110 / 0.090) pH = 4.76 + log₁₀(1.222) pH = 4.76 + 0.087 pH = 4.85

The pH changed by 0.09 units upward. Notice the symmetry: adding the same number of moles of acid or base produces the same magnitude of pH change (but in opposite directions) when the buffer starts with equal amounts of HA and A⁻.

Worked Example: Comparing Buffered vs Unbuffered

0.010 mol of NaOH is added to 1.0 dm³ of pure water. What is the pH change?

Before: pH = 7.00 After: [OH⁻] = 0.010 mol dm⁻³, [H⁺] = 1.00 × 10⁻¹², pH = 12.00

pH change = 5.00 units!

Compare this to the buffer above where the same addition caused only a 0.09 unit change. This dramatically illustrates the power of buffer solutions.

Buffer Capacity

Buffer capacity is the amount of acid or base a buffer can absorb before its pH changes significantly. It depends on:

The total concentration of buffer components — more concentrated buffers have greater capacity
The ratio of [A⁻] to [HA] — buffers work best when the ratio is close to 1 (i.e. when pH ≈ pKa)

A buffer is considered effective when the pH is within approximately 1 unit of the pKa:

Effective buffer range: pKa ± 1

Outside this range, one component becomes too depleted to absorb further additions.

Ratio [A⁻]/[HA]	pH relative to pKa	Capacity for added acid	Capacity for added base
10:1	pKa + 1	High (lots of A⁻)	Low (little HA)
1:1	pKa	Equal in both directions	Equal in both directions
1:10	pKa - 1	Low (little A⁻)	High (lots of HA)

Why does a 1:1 ratio give the best overall capacity? Because the buffer has equal reservoirs of both HA and A⁻, it can absorb equal amounts of added acid or base before either component is exhausted.

What Happens When a Buffer Is Overwhelmed

If you add more moles of acid than there are moles of A⁻, the conjugate base is completely consumed. The excess H⁺ has nothing to react with, and the pH drops sharply. Similarly, adding more base than there are moles of HA overwhelms the buffer.

Worked example: A buffer has 0.050 mol A⁻ and 0.050 mol HA. If 0.060 mol HCl is added:

0.050 mol of H⁺ reacts with all the A⁻
Excess H⁺ = 0.060 - 0.050 = 0.010 mol
The buffer is overwhelmed and the solution behaves like an acidic solution

Biological Buffers: The Blood Carbonate System

The most important biological buffer is the carbonate buffer system in blood:

CO₂(aq) + H₂O(l) ⇌ H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)

The weak acid is effectively dissolved CO₂/H₂CO₃, and the conjugate base is HCO₃⁻ (hydrogencarbonate).

Blood pH is maintained between 7.35 and 7.45. If pH falls below 7.35 (acidosis), the body responds by:

Increasing breathing rate to expel CO₂ (shifting the equilibrium left, reducing [H⁺])
The kidneys excrete more H⁺

If pH rises above 7.45 (alkalosis), the opposite responses occur.

The beauty of this system is that the body can regulate one of the buffer components (CO₂) through breathing — an extraordinary feedback mechanism that keeps pH within the narrow range compatible with life.

Other biological buffers: Proteins and the phosphate system (H₂PO₄⁻/HPO₄²⁻, pKa = 7.2) also contribute to intracellular pH regulation.

Industrial Uses of Buffers

Application	Buffer system	Target pH
Pharmaceutical manufacturing	Various	Drug-specific stability
Food production (jams, drinks)	Citric acid/citrate	3–4
Cell culture and molecular biology	PBS (phosphate-buffered saline)	7.4
Water treatment	Carbonate/bicarbonate	6.5–8.5
Electroplating	Various	Process-specific
Shampoo formulation	Citrate buffers	5–6

Comparing Buffered and Unbuffered Solutions

Property	Buffered Solution	Unbuffered Solution
Adding small amount of acid	pH barely changes	pH drops significantly
Adding small amount of base	pH barely changes	pH rises significantly
Contains	Weak acid + conjugate base	May be pure water or a salt solution
Capacity	Limited — can be overwhelmed by large additions	No resistance at all
pH depends on	Ratio [A⁻]/[HA] and pKa	Only on [H⁺] present

Summary

Buffer action depends on having two reservoirs: the weak acid to neutralise added base, and the conjugate base to neutralise added acid. Le Chatelier's principle provides the conceptual framework. Buffer capacity depends on concentration and on keeping the component ratio close to 1:1. Remember: a buffer minimises pH changes but cannot prevent them entirely, and it can be overwhelmed by large additions.

A-Level Deep Dive: How Buffers Resist pH Changes

Spec mapping

Edexcel 9CH0 specification, Topic 12: Acid-base equilibria, sub-strand 12.4 requires you to explain qualitatively how a buffer resists changes in pH on addition of small amounts of acid or base, using Le Chatelier's principle and the dissociation equilibrium of the weak acid; perform quantitative calculations of buffer pH before and after small additions of strong acid or strong base; compare buffered vs unbuffered systems quantitatively (refer to the official specification document for exact wording). The Le Chatelier explanation is mechanistic; the quantitative comparison is examined as 6–8 mark stems on Paper 1 and Paper 3.

Worked example with full mark scheme

Question (8 marks):

(a) A buffer is made from 50.0 cm³ of 0.10 mol dm⁻³ ethanoic acid (pKa = 4.76) and 50.0 cm³ of 0.10 mol dm⁻³ sodium ethanoate. Calculate the pH of the buffer. (2)

(b) 5.00 cm³ of 0.10 mol dm⁻³ HCl is added to the buffer in (a). Calculate the new pH. (4)

(c) 5.00 cm³ of 0.10 mol dm⁻³ HCl is added to 100 cm³ of pure water at 298 K (no buffer). Calculate the new pH. Comment on the difference. (2)

Solution with mark scheme:

(a) Equal moles of HA (5.00 × 10⁻³ mol) and A⁻ (5.00 × 10⁻³ mol), so [A⁻]/[HA] = 1.

pH = pKa + log₁₀(1) = 4.76.

M1 — recognises equimolar = pKa. A1 — pH = 4.76.

(b) Moles HCl added = 0.005 × 0.10 = 5.00 × 10⁻⁴ mol.

The added H⁺ reacts with the conjugate base: A⁻ + H⁺ → HA.

After reaction:

Moles HA = 5.00 × 10⁻³ + 5.00 × 10⁻⁴ = 5.50 × 10⁻³ mol.
Moles A⁻ = 5.00 × 10⁻³ − 5.00 × 10⁻⁴ = 4.50 × 10⁻³ mol.

(Total volume now 105 cm³, but cancels in ratio.)

[A⁻]/[HA] = 4.50 / 5.50 = 0.818.

pH = 4.76 + log₁₀(0.818) = 4.76 + (−0.087) = 4.67.

M1 — moles of HCl. M1 — moles of HA increased and A⁻ decreased by added H⁺. M1 — Henderson-Hasselbalch with new ratio. A1 — pH = 4.67 (so the buffer changed by only 0.09 pH units).

[H⁺] = 5.00 × 10⁻⁴ / 0.105 = 4.76 × 10⁻³ mol dm⁻³.

pH = −log₁₀(4.76 × 10⁻³) = 2.32.

Pure water is initially pH 7. Adding the same HCl to pure water changes pH by 7 − 2.32 = 4.68 units. The buffer changed pH by only 0.09 units — about 50× less change on the linear pH scale, equivalent to a [H⁺] change buffered by ≈ 50,000× less than in unbuffered water.

M1 — pH after acid addition to water. A1 — comparison with quantitative ratio.

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

Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks): Explain, using the equilibrium HA ⇌ H⁺ + A⁻ and Le Chatelier's principle, how an acidic buffer responds to (a) addition of a small amount of strong acid; (b) addition of a small amount of strong base; (c) moderate dilution with water. (6)

Mark scheme decomposition by AO:

(a)

B1 (AO1.2) — added H⁺ reacts with A⁻ via A⁻ + H⁺ → HA, removing most of the added H⁺ from solution.
B1 (AO2.1) — by Le Chatelier, the equilibrium HA ⇌ H⁺ + A⁻ shifts to the left in response to added H⁺, regenerating HA and reducing [H⁺] back toward its original value.

(b)

B1 (AO1.2) — added OH⁻ reacts with the weak acid HA via HA + OH⁻ → A⁻ + H₂O, consuming most of the added OH⁻.
B1 (AO2.1) — by Le Chatelier, the equilibrium shifts to the right (more HA dissociates) to replace the H⁺ neutralised by OH⁻.

(c)

B1 (AO2.1) — dilution decreases [H⁺] proportionally to volume, but also shifts the dissociation equilibrium to the right (Le Chatelier on a decrease in concentrations).
B1 (AO3.1) — the two effects nearly cancel: the [A⁻]/[HA] ratio is unchanged on dilution (both species diluted equally), so pH = pKa + log([A⁻]/[HA]) is unchanged. Buffers are therefore robust to moderate dilution.

Total: 6 marks all AO1/AO2/AO3. Edexcel uses Le Chatelier reasoning explicitly — bare statements like "the buffer absorbs the added acid" without mechanism do not earn full marks.

Synoptic links

Connects to:

Topic 11 — Equilibria: Le Chatelier's principle is the central tool. Buffer behaviour is the canonical application of "shift in equilibrium in response to a change in concentration" introduced in Year 1.
Topic 12.4 — Buffer composition (Lesson 5): Lesson 6 extends Lesson 5's static buffer pH calculation to dynamic response. The Henderson-Hasselbalch equation handles both regimes.