Buffer Solutions: How They Work

Learning Objectives (OCR Spec 5.1.3)

By the end of this lesson you should be able to:

• Define a buffer solution and state its purpose
• Describe the composition of an acidic buffer
• Explain how a buffer maintains near-constant pH when small amounts of acid or base are added
• Write the equilibria involved and use Le Chatelier to explain buffer action
• Recognise real-world examples, including blood buffering by HCO3-/H2CO3

What is a Buffer?

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added, or when it is diluted. Buffers do not prevent pH change completely - they limit it. When a buffer is "overwhelmed" (large quantities of H+ or OH- added), its capacity is exceeded and the pH changes sharply.

Buffers are essential wherever a biological or chemical process must occur at a specific pH:

• Blood is buffered at pH 7.35-7.45 by HCO3-/H2CO3, HPO4^2-/H2PO4- and proteins.
• Cell cytoplasm is buffered around pH 7.0-7.2.
• Shampoos and conditioners are buffered to match skin pH.
• Industrial fermentations use buffers to optimise enzyme activity.
• Titrations in analytical chemistry rely on buffered calibration solutions.

Composition of an Acidic Buffer

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

1. A weak acid HA (to react with added base OH-).
2. The conjugate base A- of that weak acid (usually supplied as a salt - to react with added acid H+).

Typical examples:

Weak acid	Salt providing A-	Buffer pH range
CH3COOH	CH3COONa (sodium ethanoate)	~3.7 to 5.7
HCOOH	HCOONa	~2.8 to 4.8
H2CO3	NaHCO3	~5.3 to 7.3
H2PO4-	HPO4^2- (mixture of salts)	~6.2 to 8.2

A basic buffer (pH > 7) consists of a weak base and its conjugate acid, e.g. NH3 / NH4Cl. OCR A-Level focuses primarily on acidic buffers.

Preparing an Acidic Buffer

There are two common methods:

Method 1: Mix a Weak Acid with its Salt

Dissolve a measured amount of the weak acid and its salt in water. E.g. 0.100 mol dm-3 CH3COOH + 0.100 mol dm-3 CH3COONa.

Method 2: Partial Neutralisation of a Weak Acid

Start with a weak acid and add less than the stoichiometric amount of strong base. The base neutralises some of the weak acid, converting it to its conjugate base while leaving unreacted weak acid behind. For example, add 0.50 mol NaOH to 1.00 mol CH3COOH in water; the product is a 1:1 buffer of CH3COOH / CH3COO-.

A particularly elegant case is half-neutralisation: when exactly half the weak acid has been converted to its salt, [HA] = [A-] and pH = pKa (see lesson 8 for the calculation).

The Buffering Equilibrium

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

CH3COOH(aq) <=> H+(aq) + CH3COO-(aq)

The weak acid provides a reservoir of undissociated HA (ready to neutralise added OH-), and the salt provides a reservoir of A- (ready to neutralise added H+). The pH is set by the ratio [HA]:[A-] and remains almost constant even as small quantities of acid or base are added.

graph TD
  A[Added H+] -->|reacts with A-| B[Forms HA]
  C[Added OH-] -->|reacts with HA| D[Forms A- and H2O]
  E[[H+] stays ~ constant]
  B --> E
  D --> E

How a Buffer Responds to Added Acid (H+)

Suppose a small amount of strong acid is added to a CH3COOH / CH3COO- buffer. The added H+ reacts with the conjugate base:

CH3COO-(aq) + H+(aq) -> CH3COOH(aq)

• [A-] decreases slightly (converted to HA).
• [HA] increases slightly.
• [H+] increases only very slightly because it is mopped up by the much larger reservoir of A-.

In Le Chatelier terms: the added H+ shifts the equilibrium to the left, consuming A- and regenerating HA. The position of the ethanoic acid equilibrium moves to restore [H+] close to its original value.

How a Buffer Responds to Added Base (OH-)

If a small amount of strong base is added instead, the OH- reacts with the weak acid:

CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)

• [HA] decreases slightly.
• [A-] increases slightly.
• [OH-] is removed, so [H+] is only very slightly raised by the equilibrium shift.

Again by Le Chatelier, the removal of H+ by OH- shifts the ethanoic acid equilibrium to the right, and HA dissociates further to replace some of the lost H+. The pH rises only very slightly.

Why It Works: The Ratio [HA] / [A-] Barely Changes