Buffers

Buffers

12 min read Updated Mar 26, 2026

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers are everywhere in biology - your blood, every cell in your body, and every biochemistry experiment relies on them. On the MCAT, buffers are one of the most frequently tested topics in all of general chemistry.

What Makes a Buffer

A buffer requires two components:

  1. A weak acid (HA) - to neutralize any added base
  2. Its conjugate base (A⁻) - to neutralize any added acid

Both must be present in significant amounts. Common buffer systems:

  • Acetic acid / sodium acetate (CH₃COOH / CH₃COO⁻)
  • Carbonic acid / bicarbonate (H₂CO₃ / HCO₃⁻) - the blood buffer
  • Dihydrogen phosphate / hydrogen phosphate (H₂PO₄⁻ / HPO₄²⁻) - intracellular buffer
  • Ammonia / ammonium (NH₃ / NH₄⁺)

How a Buffer Works - The Mechanism

When acid (H⁺) is added:

A⁻ + H⁺ → HA

The conjugate base “soaks up” the added protons, converting to the weak acid. [A⁻] decreases slightly, [HA] increases slightly, but pH barely changes.

When base (OH⁻) is added:

HA + OH⁻ → A⁻ + H₂O

The weak acid neutralizes the added hydroxide, converting to the conjugate base. [HA] decreases slightly, [A⁻] increases slightly, but pH barely changes.

Weak acid titration curve highlighting the buffer region where pH changes slowly and the pKa at the midpoint
The buffer region on a weak acid titration curve. The flat, gently sloping section before the steep rise is the buffer region, where added base causes only small pH changes. The pKa (green dashed line) marks the midpoint of this region, where [HA] = [A⁻]. Source: OpenStax.

The Henderson-Hasselbalch Equation

This is the master equation for buffer chemistry:

Key Implications of Henderson-Hasselbalch

Condition[A⁻] vs [HA]log([A⁻]/[HA])pH vs pKa
[A⁻] = [HA]Equal0pH = pKa
[A⁻] > [HA]More base formPositivepH > pKa
[A⁻] < [HA]More acid formNegativepH < pKa
[A⁻] = 10[HA]10x more base+1pH = pKa + 1
[HA] = 10[A⁻]10x more acid-1pH = pKa - 1

Making a Buffer at a Specific pH

To prepare a buffer at a desired pH:

  1. Choose a weak acid whose pKa is close to the desired pH (within ±1)
  2. Use Henderson-Hasselbalch to calculate the required ratio [A⁻]/[HA]
  3. Mix the weak acid and its conjugate base (usually as a sodium or potassium salt) in that ratio

Example: You want a buffer at pH 5.00 using acetic acid (pKa = 4.74).

pH = pKa + log([A⁻]/[HA])

5.00 = 4.74 + log([A⁻]/[HA])

log([A⁻]/[HA]) = 0.26

[A]/[HA]=100.261.8[\text{A}^-]/[\text{HA}] = 10^{0.26} \approx 1.8

You need about 1.8 times as much acetate as acetic acid.

The Bicarbonate Buffer System

The most important buffer in human physiology - and the foundation for understanding acid-base disorders in biology:

CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻

At blood pH 7.40, using pKa₁ = 6.10 for the CO₂/HCO₃⁻ system:

pH = 6.10 + log([HCO₃⁻]/[CO₂])

7.40 = 6.10 + log([HCO₃⁻]/[CO₂])

log([HCO₃⁻]/[CO₂]) = 1.30

[HCO₃⁻]/[CO₂] = 20

Normal blood has about 20 times more bicarbonate than dissolved CO₂. This ratio maintains the pH at 7.40.

Using Henderson-Hasselbalch During a Titration

Henderson-Hasselbalch works at any point during a weak acid-strong base titration (except the initial point and the equivalence point):

  1. Before any base is added: Use Ka and ICE table (no A⁻ yet)
  2. Between initial and equivalence: Use Henderson-Hasselbalch (both HA and A⁻ present)
  3. At half-equivalence: pH = pKa (simplest case)
  4. At equivalence: Use Kb of the conjugate base (all HA converted to A⁻)
  5. Past equivalence: Excess strong base determines pH
A buffer contains 0.30 M acetic acid and 0.30 M sodium acetate (pKa = 4.74). What is the pH? What happens to the pH if 0.01 mol of HCl is added to 1 L of this buffer?
Click to reveal answer
Initial pH = 4.74. [A⁻] = [HA], so log(1) = 0 and pH = pKa = 4.74. After adding 0.01 mol HCl: the H⁺ converts 0.01 mol of A⁻ to HA. New [A⁻] = 0.29 M, new [HA] = 0.31 M. pH = 4.74 + log(0.290.31\frac{0.29}{0.31}) = 4.74 + log(0.935) = 4.74 - 0.03 = 4.71. The pH dropped by only 0.03 units. Without the buffer, 0.01 mol HCl in 1 L of water would give pH = 2.0 - a drop of nearly 5 pH units.
Which of the following would make a good buffer at pH 7.2? (A) HCl / NaCl (B) CH₃COOH / CH₃COONa (pKa 4.74) (C) H₂PO₄⁻ / HPO₄²⁻ (pKa 7.21) (D) NH₃ / NH₄Cl (pKa 9.25)
Click to reveal answer
C) H₂PO₄⁻ / HPO₄²⁻. A buffer works best when pH is within ±1 of pKa. The phosphate system has pKa₂ = 7.21, which is nearly identical to the target pH 7.2. Choice A fails because HCl is a strong acid (no equilibrium). Choice B has pKa = 4.74 (too far from 7.2). Choice D has pKa = 9.25 (also too far). This is why the phosphate buffer is used in biological experiments at physiological pH.