AP Chemistry Henderson-Hasselbalch Equation

Buffer pH, solving for a target ratio, and buffer preparation, three full worked examples.

The Henderson-Hasselbalch equation is the standard tool for finding a buffer solution's pH. The formula itself is short, the actual tested skill is applying it correctly across a few different problem shapes, not just plugging concentrations straight in.

The Henderson-Hasselbalch Formula

pH = pK_a + log([A⁻]/[HA]), where [A⁻] is the conjugate base and [HA] is the weak acid. It only applies to a real buffer, a solution containing significant amounts of both. When [A⁻] = [HA], the ratio is 1, log(1) = 0, and pH = pK_a exactly, the point of maximum buffering strength.

Worked Example 1: Finding a Buffer's pH

A buffer contains 0.25 M formic acid (HCOOH, pK_a = 3.75) and 0.40 M sodium formate (HCOO⁻). What is the pH?

pH = pK_a + log([A⁻]/[HA]) = 3.75 + log(0.40/0.25) = 3.75 + log(1.6) = 3.75 + 0.20 = 3.95

Worked Example 2: Finding the Acid-Base Ratio for a Target pH

Using acetic acid (pK_a = 4.74), what ratio of [A⁻] to [HA] is needed to make a buffer with pH = 5.00?

5.00 = 4.74 + log([A⁻]/[HA])
log([A⁻]/[HA]) = 0.26
[A⁻]/[HA] = 10^0.26 = 1.82

A ratio above 1 makes sense: the target pH (5.00) is above the acid's pK_a (4.74), so the solution needs more conjugate base than acid.

Worked Example 3: Finding the pH of a Buffer Made From a Reaction

100. mL of 0.50 M acetic acid (pK_a = 4.74) is mixed with 30.0 mL of 0.50 M NaOH. What is the resulting solution's pH?

This is a stoichiometry problem first, then a Henderson-Hasselbalch problem:
mol HA (initial) = (0.100 L)(0.50 M) = 0.0500 mol
mol OH⁻ added = (0.0300 L)(0.50 M) = 0.0150 mol
HA + OH⁻ → A⁻ + H₂O consumes HA 1:1 with the added base:
mol HA remaining = 0.0500 − 0.0150 = 0.0350 mol
mol A⁻ formed = 0.0150 mol

Since both are in the same total volume, moles can substitute directly for concentration in the ratio:
pH = 4.74 + log(0.0150/0.0350) = 4.74 + (−0.37) = 4.37

This is a different skill from calculating a pH change after adding acid or base to an already-existing buffer, that specific numerical calculation is excluded. Here, the buffer doesn't exist yet until the reaction happens, so finding its resulting pH is a fully tested combination of stoichiometry and Henderson-Hasselbalch.

Common Henderson-Hasselbalch Mistakes

Flipping the ratio. It's [A⁻] (conjugate base) over [HA] (weak acid), not the other way around.
Using Henderson-Hasselbalch when there's no real buffer. The equation doesn't apply at the equivalence point of a titration, or to a solution of just the weak acid with no conjugate base present.
Trying to calculate an exact pH change after adding acid or base to an existing buffer. That numerical calculation is explicitly excluded, only qualitative reasoning about why the change is small is tested for that specific scenario.
Forgetting to do stoichiometry first in a preparation problem. If a buffer forms from a reaction, find the moles of acid and conjugate base actually present after the reaction completes before applying Henderson-Hasselbalch.

Frequently Asked Questions

What is the Henderson-Hasselbalch equation?

pH = pKa + log([A⁻]/[HA]), where [A⁻] is the concentration (or moles) of the conjugate base and [HA] is the concentration (or moles) of the weak acid. It calculates the pH of a buffer solution, a mixture containing significant amounts of both a weak acid and its conjugate base.

Do I need to know how to derive the Henderson-Hasselbalch equation?

No. Deriving the equation is explicitly excluded from AP Chemistry. You need to apply pH = pKa + log([A⁻]/[HA]) correctly, not prove where it comes from.

Can I use moles instead of concentration in the equation?

Yes. Since both the acid and conjugate base are dissolved in the same total volume, that volume cancels out of the ratio, moles of A⁻ over moles of HA gives the same answer as concentration of A⁻ over concentration of HA. This is especially useful in buffer-preparation problems where you calculate moles first.

Does AP Chemistry ask you to calculate the new pH after adding acid or base to a buffer?

No, that specific numerical calculation, the pH change after adding a small amount of strong acid or base to an already-existing buffer, is explicitly excluded. You only need to explain qualitatively why the change is small. Calculating a buffer's pH when it's first prepared, including through a neutralization reaction, is a different, fully tested skill, covered in this article's third example.

What does it mean when pH equals pKa?

When [A⁻] equals [HA], their ratio is 1, and log(1) = 0, so pH = pKa exactly. This is the point of maximum buffering capacity, and it's why an effective buffer is chosen using an acid whose pKa is close to the target pH.

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