Henderson-Hasselbalch Equation
Calculate the current pH of an established buffer from a justified base/acid ratio.
AP Chemistry Topic 8.9 · CED printed p. 157 / PDF p. 164. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Conjugate-pair ratio
- Logarithms
- Mixture composition
For an established weak conjugate-pair buffer, . This Henderson–Hasselbalch relation comes from the pair’s equilibrium expression. It tells you the current pH from the acid identity and the base/acid ratio. Equal amounts give pH = pKa; more base gives pH above pKa; more acid gives pH below pKa. A tenfold ratio change corresponds to one pH unit. Both components must be present in meaningful amounts; the formula is not a shortcut for a strong-acid-only solution or an equivalence-point solution with essentially only one weak form.
For ordinary dilute buffers, use the component concentrations after any preparation reaction, assuming their subsequent equilibrium changes are small. A common final volume cancels from the ratio, so moles of the two buffer forms can replace concentrations when both are in the same mixture. Uniform dilution leaves the ratio approximately unchanged, but it reduces capacity. To choose a target composition, use [A−]/[HA] = 10^(pH − pKa), then check that both forms remain appreciable. The AP course assesses the current buffer pH and its qualitative response; formal derivation and calculation of the change in pH after adding acid/base to a buffer are excluded. You can explain a response with the changing ratio without calculating that excluded pH difference.
Let's work through it
- A buffer pair has pKa = 9.20. In one final solution, nHB+ = 0.0200 mol and nB = 0.0300 mol.
- Treat HB+ as the acid and B as its base. The common volume cancels, so pH = 9.20 + log(0.0300/0.0200).
- The ratio is 1.50; log 1.50 = 0.176. pH = 9.38, with two decimal places limited by pKa.
- B is more abundant, so pH above 9.20 makes sense. Reversing the ratio would fail this chemical check.
Make the picture move
Interactive concept model
The species change across a titration curve.
Predict first. Visit 0, 12.5, 25.0, and 37.5 mL. Identify the species controlling pH at each point. Is equivalence necessarily pH 7?
HA + OH⁻ → A⁻ + H₂O; Kₐ = [H⁺][A⁻]/[HA]; K_w = [H⁺][OH⁻]
- Ideal weak-acid titration
- Labeled selected or reference state
On a small screen, scroll inside the graph or diagram to read its labels. The values and explanation also appear below.
At 12.5 mL, pH = 4.74. Half-equivalence: pH ≈ pKₐ. [HA] = 0.0333 M and [A⁻] = 0.0334 M.
Read the data and comparisons
| Added base (mL) | pH | [HA] (mol L⁻¹) | [A⁻] (mol L⁻¹) |
|---|---|---|---|
| 0.0 | 2.87 | 0.0987 | 0.00134 |
| 12.5 | 4.74 | 0.0333 | 0.0334 |
| 25.0 | 8.72 | 5.24 × 10^-6 | 0.05 |
| 37.5 | 12.30 | 1.10 × 10^-9 | 0.04 |
| 50.0 | 12.52 | 5.50 × 10^-10 | 0.0333 |
Macroscopic / data. The pH curve is a computed data representation, with a changing total liquid volume.
Particle view. HA, A⁻, H⁺, OH⁻, and spectator Na⁺ satisfy matter and charge conservation. Equivalence does not remove the hydrolysis equilibrium.
Symbolic view. Total acid-species concentration = 0.0667 M; [H⁺] = 1.82 × 10^-5 M; [OH⁻] = 5.50 × 10^-10 M.
Why it works. The initial 25.0 mL of 0.100 M HA contains 0.00250 mol of acid. Equivalence takes 25.0 mL of 0.100 M base regardless of pKₐ. At half-equivalence, comparable HA and A⁻ give pH approximately pKₐ. Within the buffer region, pH = pKₐ + log₁₀([A⁻]/[HA]) follows by rearranging Kₐ; estimating this ratio from neutralization requires appreciable amounts of both species. At equivalence A⁻ reacts with water, so a weak-acid titration is basic. Beyond equivalence excess OH⁻ dominates.
Conditions and limits. Monoprotic hypothetical HA and a fully dissociated strong base with Na⁺ as spectator; 25 °C, K_w = 1.0 × 10⁻¹⁴, ideal dilute activities approximated by concentrations divided by a 1 mol L⁻¹ standard state, additive volumes, no other equilibria. The curve solves acid mass balance, charge balance, Kₐ, and K_w together at every volume; it does not misuse Henderson–Hasselbalch at endpoints. The numerical solver is an implementation detail, not required AP work. pH is displayed to two decimal places.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
Which form is on top in the logarithm?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
Which form is on top in the logarithm?
Worked answer and reason
The conjugate base; the conjugate acid is underneath.
More base should raise pH relative to pKa.
Retrieval 2
Does equal dilution substantially change buffer pH?
Worked answer and reason
In the ideal model it preserves the ratio and approximately preserves pH, while lowering capacity.
The ratio sets pH; reservoir size sets capacity.
Catch the tempting mistake
- Use Henderson–Hasselbalch at equivalence with no appreciable HA.
- At equivalence, solve the remaining conjugate species’ weak equilibrium instead.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 8 problem 13 · ap application · symbolic
Scientific calculator workspace
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Scientific calculator
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For the teacher
- Require a dominance check against pH − pKa. Distinguish calculating an established composition from the AP-excluded buffer pH-change computation.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 8.9.A
- Identify the pH of a buffer solution based on the identity and concentrations of the conjugate acid-base pair used to create the buffer.
- 8.9.A.1
- The pH of the buffer is related to the pK_a of the acid and the concentration ratio of the conjugate acid-base pair. This relation is a consequence of the equilibrium expression associated with the dissociation of a weak acid, and is described by the Henderson-Hasselbalch equation. Adding small amounts of acid or base to a buffered solution does not significantly change the ratio of [A−]/[HA] and thus does not significantly change the solution pH. The change in pH on addition of acid or base to a buffered solution is therefore much less than it would have been in the absence of the buffer. EQN: pH = pK_a + log([A^−]/[HA])
- Computation of the change in pH resulting from the addition of an acid or a base to a buffer will not be assessed on the AP Exam.
- Derivation of the Henderson-Hasselbalch equation will not be assessed on the AP Exam.
Science references for this teaching
- AP Chemistry Course and Exam Description, effective Fall 2024 — Required topic-numbered learning objectives and essential knowledge, all continuations and AP exclusions; supplied PDF SHA256 b5dfe8677ef3d88c613865d2e2a3e8d6125d652e2b24c71ef1e8ce4e011094f0.
- OpenStax Chemistry 2e §14.6, Buffers — Conjugate-pair buffer reactions, composition ratio, and capacity.
Glossary and deeper reading · 5 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Mixture
A mixture contains two or more substances whose proportions can vary. Its components retain their chemical identities until a chemical reaction changes them. For example, water and ethanol can form mixtures with many different compositions.
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Buffer
A buffer contains appreciable amounts of a weak acid and its conjugate base, or the corresponding weak-base pair. Those partners consume small added amounts of strong base or acid, limiting the pH change within the buffer's capacity. For example, an acetic acid–acetate mixture can resist modest additions of acid or base.
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pH
pH is the negative base-ten logarithm of hydronium activity. In suitable dilute models, activity is approximated by molar concentration relative to the standard concentration. For example, lowering hydronium concentration by a factor of ten raises pH by one unit.
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Dilution
Dilution lowers solute concentration by adding solvent while conserving the solute amount when no reaction or loss occurs. Concentration calculations use the final solution volume rather than simply the volume of solvent added. For example, increasing a solution's volume from 100 to 200 milliliters halves its concentration if its solute amount is unchanged.
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Chemical equilibrium
Chemical equilibrium is a dynamic state in which forward and reverse reaction rates are equal. Macroscopic composition remains constant in a closed system under fixed conditions, but reactant and product concentrations need not be equal. For example, a reversible reaction can reach equilibrium with much more reactant than product.