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Henderson-Hasselbalch Equation

AP Course

AP Chemistry

Updated for the AP Chemistry framework effective fall 2024, with particle models, quantitative reasoning, labs, quizzes, and practice problems.

Build general chemistry fluency through atomic structure, bonding, reactions, kinetics, equilibrium, acids and bases, thermodynamics, and electrochemistry.

AP Chemistry · Unit 8 · Topic 8.9

Henderson-Hasselbalch Equation

This equation is not a separate law of buffers. It is the acid equilibrium written in a form that makes a concentration ratio easy to interpret.

Derive it from the equilibrium expression

Start with \(K_a=[H_3O^+][A^-]/[HA]\) in the dilute approximation. Rearranging gives \([H_3O^+]=K_a[HA]/[A^-]\). Taking negative base-ten logarithms yields

\[\mathrm{pH}=\mathrm{p}K_a+\log\frac{[A^-]}{[HA]}.\]

The thermodynamic version uses activities. The ratio is base over acid, so a larger base fraction should increase the predicted pH; this is a useful sign check.

Understand the practical approximation

The rearranged equilibrium equation uses equilibrium quantities. In ordinary buffer problems, we approximate those with the amounts after any strong acid–base reaction because the subsequent weak ionization changes them little. If the buffer is extremely dilute or one component nearly exhausted, this substitution may be poor even though the equilibrium relation itself remains sound.

Plan an original target mixture

A hypothetical conjugate pair has pKa = 4.80. Design 250.0 mL with total buffer concentration [HA] + [A⁻] = 0.100 M and target pH 5.10. First calculate \(r=10^{5.10-4.80}=1.995\). The total pair amount is 25.0 mmol. Solving \(n_{A^-}=rn_{HA}\) and \(n_{A^-}+n_{HA}=25.0\) gives approximately 16.65 mmol A⁻ and 8.35 mmol HA.

Keep the design ledger readable

Original buffer design, rounded preparation targets
QuantityValueCheck
HA amount8.35 mmolAcid component
A⁻ amount16.65 mmolBase component
Total amount25.0 mmol0.100 M in 250.0 mL
Base/acid ratio1.994pH ≈ 5.10

Use moles only when the volume cancels

Within one mixed solution, both species share the same final volume, so their concentration ratio equals their mole ratio. This does not mean volume never matters. Volume is required to check total concentration, capacity per liter, and whether the equilibrium approximation is valid. Do not divide moles from separate, unmixed containers and call it a solution ratio.

Correct for a strong-acid addition first

If 2.00 mmol HCl is added to the designed buffer, it consumes A⁻ and produces HA. The new ratio is \(14.65/10.35\), giving pH ≈ 4.95. Inserting the original ratio would ignore the reaction; inserting the added HCl concentration directly as equilibrium hydronium would ignore the buffer's response.

Validate the preparation

Prepare with appropriate teacher-approved materials, bring the solution to its final volume, mix, and measure pH at the intended temperature. Differences from the calculation may reflect activity coefficients, reagent hydration, concentration uncertainty, or electrode calibration. Adjusting a real buffer requires recording what was added so the final composition remains known.

Practice with answers

  1. For pKa 6.00 and A⁻/HA = 0.10, find pH. 5.00.
  2. To reach pH = pKa + 0.60, what ratio is needed? About 3.98 base to 1 acid.
  3. Can you use this ordinary buffer calculation after all HA is consumed? No; choose the appropriate hydrolysis or excess-strong-base model.

Factual reference: OpenStax Chemistry 2e, 14.6. The derivation's presentation, design target, preparation ledger, and practice are original.

Official curriculum reference: College Board AP Chemistry course page. The explanation and worked example are independently written for this study site.