The Henderson–Hasselbalch Equation with Worked Examples

The Henderson–Hasselbalch equation is the single most useful formula for buffers. It tells you the pH of a mixture of a weak acid and its conjugate base in one line, and it works just as well backwards: tell it the pH you want, and it tells you the recipe.

It’s used by biochemists preparing enzyme solutions, by doctors interpreting blood tests, by pharmacologists predicting how drugs behave, and by every chemistry student who meets buffers. This guide derives it, shows how to use it, and explains where it stops being reliable.

The equation

pH = pKa + log₁₀([A⁻] ÷ [HA])

where:

• HA is the weak acid
• A⁻ is its conjugate base (usually supplied as a salt, such as sodium ethanoate)
• pKa is −log Ka for the weak acid

For a weak base buffer (such as ammonia with ammonium chloride), you can use the same equation with the pKa of the conjugate acid: pH = pKa(NH₄⁺) + log([NH₃] ÷ [NH₄⁺]).

Where it comes from

Start from the acid dissociation constant:

Ka = [H⁺][A⁻] ÷ [HA]

Rearrange for [H⁺]:

[H⁺] = Ka × [HA] ÷ [A⁻]

Take −log of both sides:

−log[H⁺] = −log Ka − log([HA] ÷ [A⁻])

pH = pKa + log([A⁻] ÷ [HA])

(The sign flips because −log(x ÷ y) = +log(y ÷ x).)

That’s it: no new chemistry, just the Ka expression in logarithmic form.

What it tells you at a glance

• If [A⁻] = [HA], the log term is log 1 = 0, so pH = pKa. A buffer with equal amounts of acid and conjugate base has a pH equal to the acid’s pKa.
• If there’s more A⁻ than HA, the log term is positive and the pH is above pKa.
• If there’s more HA, the pH is below pKa.
• A tenfold change in the ratio shifts the pH by exactly one unit.

[A⁻] : [HA]	log(ratio)	pH
1 : 10	−1	pKa − 1
1 : 2	−0.30	pKa − 0.30
1 : 1	0	pKa
2 : 1	+0.30	pKa + 0.30
10 : 1	+1	pKa + 1

A shortcut worth knowing: use moles

Because the acid and its conjugate base are in the same solution, they share the same volume. The volume cancels in the ratio, so you can use moles instead of concentrations:

pH = pKa + log(moles A⁻ ÷ moles HA)

This saves a step in almost every problem.

Worked example 1: pH of a simple buffer

A buffer contains 0.20 mol/dm³ ethanoic acid and 0.30 mol/dm³ sodium ethanoate. pKa = 4.76. Find the pH.

pH = 4.76 + log(0.30 ÷ 0.20) = 4.76 + log(1.5) = 4.76 + 0.18 = 4.94

Worked example 2: mixing two solutions

50.0 cm³ of 0.100 mol/dm³ ethanoic acid is mixed with 25.0 cm³ of 0.100 mol/dm³ sodium ethanoate. Find the pH.

• Moles HA = 0.100 × 0.0500 = 5.00 × 10⁻³ mol
• Moles A⁻ = 0.100 × 0.0250 = 2.50 × 10⁻³ mol
• pH = 4.76 + log(2.50 ÷ 5.00) = 4.76 − 0.30 = 4.46

You don’t need the total volume at all.

Worked example 3: finding the ratio for a target pH

You need a buffer at pH 5.00 using ethanoic acid and sodium ethanoate. What ratio of [A⁻] to [HA] is required?

5.00 = 4.76 + log(ratio) log(ratio) = 0.24 ratio = 10⁰·²⁴ = 1.74

So you need 1.74 times as much ethanoate as ethanoic acid (in moles).

Worked example 4: making a buffer from a solid

What mass of sodium ethanoate (CH₃COONa, M = 82.03 g/mol) must be dissolved in 500 cm³ of 0.100 mol/dm³ ethanoic acid to make a buffer of pH 5.00? Assume no volume change.

• From example 3, moles A⁻ ÷ moles HA = 1.74
• Moles HA = 0.100 × 0.500 = 0.0500 mol
• Moles A⁻ = 1.74 × 0.0500 = 0.0870 mol
• Mass = 0.0870 × 82.03 = 7.14 g

Worked example 5: a buffer made by partial neutralisation

Buffers are often made by adding some strong base to a weak acid, not enough to neutralise it all. The base converts part of the acid into its conjugate base.

25.0 cm³ of 0.100 mol/dm³ NaOH is added to 50.0 cm³ of 0.100 mol/dm³ methanoic acid (pKa 3.75). Find the pH.

• Moles HCOOH at start = 0.100 × 0.0500 = 5.00 × 10⁻³ mol
• Moles OH⁻ added = 0.100 × 0.0250 = 2.50 × 10⁻³ mol
• Reaction: HCOOH + OH⁻ → HCOO⁻ + H₂O
• After reaction: HCOOH = 2.50 × 10⁻³ mol; HCOO⁻ = 2.50 × 10⁻³ mol
• pH = 3.75 + log(1) = 3.75

This is exactly the half-equivalence point of a titration, where pH = pKa. See titration curves.

Worked example 6: a basic buffer

A buffer contains 0.15 mol/dm³ ammonia and 0.10 mol/dm³ ammonium chloride. pKa(NH₄⁺) = 9.25. Find the pH.

Here NH₄⁺ is the acid and NH₃ is the conjugate base:

pH = 9.25 + log(0.15 ÷ 0.10) = 9.25 + 0.18 = 9.43

Worked example 7: biology

Blood is buffered mainly by carbonic acid and hydrogencarbonate. At body temperature, the effective pKa is about 6.1. Normal blood has pH 7.40. What is the ratio [HCO₃⁻] : [H₂CO₃]?

7.40 = 6.1 + log(ratio) log(ratio) = 1.30 ratio = 10¹·³⁰ ≈ 20 : 1

Blood contains about twenty times as much hydrogencarbonate as carbonic acid. That lopsided ratio seems odd for a buffer, but it works because the lungs can remove CO₂ (and so H₂CO₃) very quickly. See the blood buffer system.

Worked example 8: drug ionisation

Aspirin has pKa 3.5. What fraction is ionised in the stomach (pH 1.5) and in the blood (pH 7.4)?

• Stomach: log([A⁻] ÷ [HA]) = 1.5 − 3.5 = −2, so [A⁻] : [HA] = 1 : 100. About 1% ionised, 99% neutral.
• Blood: log(ratio) = 7.4 − 3.5 = 3.9, so [A⁻] : [HA] ≈ 8000 : 1. Over 99.9% ionised.

The neutral form crosses cell membranes much more easily, which helps explain why aspirin is absorbed in the stomach lining and why it’s then trapped in ionised form once it’s in the blood.

When the equation works, and when it doesn’t

The Henderson–Hasselbalch equation assumes that the equilibrium concentrations of HA and A⁻ are the same as the amounts you added. That’s a good approximation when:

• Both HA and A⁻ are present in reasonable amounts (ratio between about 0.1 and 10).
• Concentrations are not too dilute (roughly above 0.01 mol/dm³).
• The pKa is between about 3 and 11, so that neither the acid’s own ionisation nor the base’s reaction with water shifts the amounts significantly.

It becomes unreliable for:

• Very dilute buffers, where water’s own ions matter.
• Relatively strong weak acids (pKa below about 3) at low concentration, where a significant fraction of HA ionises.
• Solutions containing only acid or only salt. Then the ratio is zero or infinite and the equation breaks. Use pH of weak acids or pH of weak bases instead.
• High ionic strength, where activities differ from concentrations. Precise work uses activity coefficients.

Common mistakes

• Flipping the ratio. It’s base over acid: [A⁻] ÷ [HA]. Sanity check: more base should mean higher pH.
• Using Ka instead of pKa.
• Forgetting the neutralisation step when strong acid or base is added. React first, then apply the equation.
• Using it for a strong acid. Strong acids don’t form buffers with their conjugate bases.

Key takeaways

• pH = pKa + log([A⁻] ÷ [HA]), derived directly from the Ka expression.
• When [A⁻] = [HA], pH = pKa; each tenfold change in ratio shifts pH by one unit.
• Moles can replace concentrations because the volume cancels.
• It works best for reasonable concentrations and ratios between 0.1 and 10.
• For what happens when acid or alkali is added to a buffer, see buffer calculations.