Henderson-Hasselbalch Equation

8.9 Henderson-Hasselbalch

Henderson-Hasselbalch Equation

Drag [HA] and [A⁻] sliders and watch pH = pKₐ + log([A⁻]/[HA]) substitute term by term across 6 weak acids. Watch the pH bead land inside the pKₐ ± 1 buffer zone.

pH = pKₐ + log(A⁻/HA)6 Weak AcidsHalf-Eq PointpKₐ ± 1 Range

Topic 8.9

Henderson–Hasselbalch Equation

Identify the pH of a buffer solution based on the identity and concentrations of the conjugate acid–base pair used to create the buffer.

The pH of a 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 for the dissociation of a weak acid and is described by the Henderson–Hasselbalch equation, printed on the AP equation sheet:

pH = pK_a + log([A⁻]/[HA])

The same EK explains the buffering mechanism quantitatively: adding small amounts of acid or base to a buffered solution does not significantly change the ratio [A⁻]/[HA], and therefore does not significantly change the pH. The change in pH on adding acid or base to a buffer is much less than it would be without the buffer.

Three readings of the equation:

[A⁻] = [HA] → log(1) = 0 → pH = pK_a. This is the half-equivalence point (8.5) and the center of the buffer's effective range.
[A⁻] > [HA] → pH > pK_a.
[A⁻] < [HA] → pH < pK_a.

A useful practical point: because both species are in the same solution, the volume cancels in the ratio. You may use moles instead of concentrations — which saves a step whenever a buffer is formed by mixing.

Designing a buffer: choose an acid whose pK_a is within about one unit of the target pH, then set the ratio to fine-tune. A buffer is effective roughly over pK_a ± 1, corresponding to ratios between 1:10 and 10:1.

Validity limits: the equation assumes both members of the pair are present in reasonable amounts and that the "x is small" approximation holds. It is not valid for a solution of a weak acid alone.

Key points

pH = pK_a + log([A⁻]/[HA]), with base over acid inside the log.
Equal concentrations → pH = pK_a.
Moles may be used instead of concentrations, since the volume cancels.
A buffer works well over roughly pK_a ± 1.

Equations

$pH = p K_{a} + lo g \frac{[ A ^{-} ]}{[ HA ]}$
on the exam sheetPrinted on the AP equation sheet. Base on top.
- $[A^{-}]$ conjugate base concentration (or moles)
- $[HA]$ weak acid concentration (or moles)

Common mistakes

Inverting the ratio. Base goes on top. Getting it backwards flips the pH about pK_a.
Using pK_b with this equation. Convert to pK_a of the conjugate acid first (pK_a = 14 − pK_b at 25 °C).
Applying it to a weak acid with no conjugate base present. That requires an ICE table.
Forgetting to use the total volume if you choose to work in concentrations rather than moles.

Worked example

(a) Calculate the pH of a buffer made by dissolving 0.30 mol of NH₄Cl in 1.0 L of 0.50 M NH₃ (K_b for NH₃ = 1.8 × 10⁻⁵). (b) A chemist needs a buffer at pH 5.00 and has acetic acid (pK_a 4.74) available. What ratio of acetate to acetic acid is required?

(a) Ammonia buffer.

Henderson–Hasselbalch uses pK_a of the conjugate acid, which here is NH₄⁺.
K_a(NH₄⁺) = K_w/K_b = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰
pK_a = −log(5.56 × 10⁻¹⁰) = 9.26

Identify the pair: NH₄⁺ is the acid (HA), NH₃ is the base (A⁻).
pH = 9.26 + log(0.50 / 0.30) = 9.26 + log(1.67) = 9.26 + 0.22
pH = 9.48

Above pK_a, as expected since there is more base than acid.

(b) Designing a pH 5.00 acetate buffer.
5.00 = 4.74 + log([A⁻]/[HA])
log([A⁻]/[HA]) = 0.26
[A⁻]/[HA] = 10^0.26 = 1.8 : 1 acetate to acetic acid

This is a good choice of buffer system: the target pH is within 1 unit of the pK_a, so the required ratio stays well inside the 1:10 to 10:1 range where buffering is effective.

Full notes for topic 8.9 →