Buffer Solutions and How They Work

Buffer Solutions and How They Work

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added, or when the solution is diluted.

Types of Buffer

Acidic buffer (pH < 7): Made from a weak acid and its conjugate base (typically as the sodium or potassium salt).

Example: Ethanoic acid (CH₃COOH) + sodium ethanoate (CH₃COONa)

Basic buffer (pH > 7): Made from a weak base and its conjugate acid (typically as the chloride salt).

Example: Ammonia (NH₃) + ammonium chloride (NH₄Cl)

How an Acidic Buffer Works

The buffer contains a large reservoir of both the weak acid (HA) and its conjugate base (A⁻):

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

The weak acid is only slightly dissociated, so HA molecules are in large excess. The salt is fully dissociated, providing a large concentration of A⁻ ions.

When a small amount of acid (H⁺) is added:

• The added H⁺ ions react with the A⁻ ions (conjugate base) from the salt:

H⁺ + A⁻ → HA

• This removes the added H⁺, preventing a significant drop in pH
• The equilibrium shifts to the left, but the pH change is very small because [A⁻] is large

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

• The added OH⁻ ions react with the HA molecules (weak acid):

OH⁻ + HA → A⁻ + H₂O

• This removes the added OH⁻, preventing a significant rise in pH
• The equilibrium shifts to the right, but [HA] is large so the pH change is very small

Key insight: The buffer works because both HA and A⁻ are present in large, roughly comparable concentrations. The weak acid neutralises added base; the conjugate base neutralises added acid.

Calculating the pH of a Buffer

For an acidic buffer, the Ka expression applies:

Ka = [H⁺][A⁻] / [HA]

Rearranging: [H⁺] = Ka × [HA] / [A⁻]

Then pH = −log₁₀[H⁺]

This can also be expressed as the Henderson-Hasselbalch equation:

pH = pKa + log₁₀([A⁻] / [HA])

Worked Example 1: Mixing Acid and Salt

A buffer is prepared by mixing 100 cm³ of 0.200 mol dm⁻³ ethanoic acid with 100 cm³ of 0.100 mol dm⁻³ sodium ethanoate. Ka for ethanoic acid = 1.74 × 10⁻⁵ mol dm⁻³.

Step 1: Calculate moles of each component (before the equilibrium — since both solutions are diluted by mixing, we need concentrations in the final volume):

• Moles CH₃COOH = 0.200 × 0.100 = 0.0200 mol
• Moles CH₃COO⁻ = 0.100 × 0.100 = 0.0100 mol
• Total volume = 200 cm³ = 0.200 dm³

Step 2: Concentrations in the mixture:

• [CH₃COOH] = 0.0200 / 0.200 = 0.100 mol dm⁻³
• [CH₃COO⁻] = 0.0100 / 0.200 = 0.0500 mol dm⁻³

Step 3: Calculate [H⁺]:

[H⁺] = Ka × [HA] / [A⁻] = (1.74 × 10⁻⁵ × 0.100) / 0.0500 = 3.48 × 10⁻⁵ mol dm⁻³

Step 4: pH = −log₁₀(3.48 × 10⁻⁵) = 4.46

Note: Since we are taking a ratio [HA]/[A⁻], we can use moles directly (the volume cancels). So [H⁺] = Ka × (0.0200/0.0100) = Ka × 2 = 3.48 × 10⁻⁵. This shortcut is very useful.

Worked Example 2: Adding Acid to a Buffer

To the buffer above (0.0200 mol HA, 0.0100 mol A⁻), 0.002 mol HCl is added. Calculate the new pH.

The added H⁺ reacts with A⁻:

• New moles A⁻ = 0.0100 − 0.002 = 0.0080 mol
• New moles HA = 0.0200 + 0.002 = 0.0220 mol

[H⁺] = Ka × (moles HA / moles A⁻) = 1.74 × 10⁻⁵ × (0.0220 / 0.0080) = 4.79 × 10⁻⁵

pH = −log₁₀(4.79 × 10⁻⁵) = 4.32

The pH dropped from 4.46 to 4.32 — a change of only 0.14 pH units, demonstrating the buffering action.

Worked Example 3: Making a Buffer with a Target pH

What ratio of CH₃COOH to CH₃COONa is needed for a buffer at pH 5.00? (Ka = 1.74 × 10⁻⁵)

[H⁺] = 10⁻⁵·⁰⁰ = 1.00 × 10⁻⁵ mol dm⁻³

[HA] / [A⁻] = [H⁺] / Ka = (1.00 × 10⁻⁵) / (1.74 × 10⁻⁵) = 0.575

So the ratio of acid to salt is 0.575 : 1, or equivalently the moles of salt should be about 1.74 times the moles of acid.

Buffer Capacity

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

• The total concentrations of HA and A⁻ — more concentrated buffers have greater capacity
• The ratio [HA]/[A⁻] — a buffer is most effective when [HA] ≈ [A⁻], i.e. when pH ≈ pKa

A buffer is generally effective within ±1 pH unit of the pKa of the weak acid.

Biological Buffers

Blood pH is maintained at ~7.40 by the carbonic acid / hydrogencarbonate buffer system:

H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)

• If blood pH drops (excess H⁺): HCO₃⁻ reacts with H⁺ to form H₂CO₃, which decomposes to CO₂ and H₂O; the CO₂ is exhaled
• If blood pH rises (excess OH⁻ or loss of H⁺): H₂CO₃ releases H⁺ to compensate

Other biological buffers include the phosphate buffer (H₂PO₄⁻ / HPO₄²⁻) inside cells and amino acid / protein buffers.

Exam Tips

• When explaining how a buffer works, you must identify BOTH the component that removes added acid (A⁻) and the component that removes added base (HA), and write the relevant equations
• In calculations, you can use the moles ratio directly instead of calculating concentrations (the volume cancels)
• Remember: pH = pKa when [HA] = [A⁻] — this is the point of maximum buffer capacity
• When making a buffer by partial neutralisation (adding some NaOH to a weak acid), the NaOH converts some HA to A⁻