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Buffers & Henderson–Hasselbalch

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What makes a buffer

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. It contains an appreciable amount of both a weak acid (HA) and its conjugate base (A⁻). When acid is added, the A⁻ neutralizes it; when base is added, the HA neutralizes it. The reservoir of each partner soaks up the shock, so the pH barely moves.

From Ka to Henderson–Hasselbalch

Rearranging Ka = [H⁺][A⁻] ÷ [HA] and taking −log of both sides converts the equilibrium into a direct pH formula. The result shows that a buffer’s pH is anchored near the acid’s pKa and nudged up or down only by the ratio of conjugate base to acid — not by the absolute amounts. That is why diluting a buffer barely changes its pH.

Henderson–Hasselbalch equation
pH = pKa + log([A⁻] / [HA])
When [A⁻] = [HA] the log term is log(1) = 0, so pH = pKa. More conjugate base raises pH; more acid lowers it.
Worked example

A buffer is 0.20 M in acetate (A⁻) and 0.10 M in acetic acid (HA). Acetic acid has pKa = 4.74. Find the pH.

  1. 1.Use pH = pKa + log([A⁻] / [HA]).
  2. 2.Ratio [A⁻]/[HA] = 0.20 / 0.10 = 2.0.
  3. 3.log(2.0) = 0.30.
  4. 4.pH = 4.74 + 0.30 = 5.04.
Answer: pH = 5.04
Tip

Because only the ratio matters, you may plug in moles instead of concentrations — the shared volume cancels in [A⁻]/[HA]. This is a real time-saver when a problem gives you moles of acid and base directly.

Checkpoint

A buffer contains equal concentrations of a weak acid (pKa = 4.74) and its conjugate base. What is its pH?

Checkpoint

A buffer is 0.10 M in conjugate base A⁻ and 0.20 M in acid HA, with pKa = 4.74. What is the pH?

On the exam

A buffer works best when pKa is within about 1 unit of the target pH, i.e. the [A⁻]/[HA] ratio stays between 1:10 and 10:1. To build a buffer for a given pH, choose a weak acid whose pKa is close to that pH.

Answer the 2 checkpoints as you read.

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