The Henderson–Hasselbalch Equation

The Henderson-Hasselbalch equation, pH = pK_a + log([A⁻] ÷ [HA]), gives the pH of a buffer from the pK_a of the weak acid and the ratio of the concentrations of its conjugate base and the acid. In brief: when [A⁻] = [HA], pH = pK_a; a ratio of 10 : 1 raises the pH one unit above the pK_a, and 1 : 10 lowers it one unit; in blood, a ratio of 20 : 1 between hydrogencarbonate and dissolved CO₂ gives pH 7.40.

The equation is the main tool for working with buffers, for designing solutions of a desired pH and for understanding how pH depends on ionization in biochemistry and pharmacology. This article derives it, explains when it is valid, shows how to use it for preparing buffers and for finding the ionization state of a drug or amino acid, includes a table of ratios, works through examples and gives some history.

How is the equation derived?

Start from K_a = [H⁺][A⁻] ÷ [HA]. Rearrange: [H⁺] = K_a × [HA] ÷ [A⁻]. Take −log of both sides: −log[H⁺] = −log K_a − log([HA] ÷ [A⁻]). Since −log of an inverse is +log, pH = pK_a + log([A⁻] ÷ [HA]). Because the ratio of concentrations in the same volume equals the ratio of moles, moles can be used.

What do different ratios mean?

Effect of the [A⁻] : [HA] ratio on pH
[A⁻] : [HA]	log(ratio)	pH
1 : 100	−2	pK_a − 2
1 : 10	−1	pK_a − 1
1 : 2	−0.30	pK_a − 0.30
1 : 1	0	pK_a
2 : 1	+0.30	pK_a + 0.30
10 : 1	+1	pK_a + 1
100 : 1	+2	pK_a + 2

For acetic acid (pKa = 4.76) the pH rises one unit for every tenfold increase in the ratio [A−] : [HA]. At a ratio of 1 : 1 (log = 0) the pH equals pKa. The shaded band, ratios from 1 : 10 to 10 : 1, is where the equation works best.

When is it valid?

The concentrations of acid and base must be large enough (at least about 100 times K_a) that the amount ionized does not change [HA] and [A⁻] appreciably.
The ratio should lie between about 0.1 and 10.
The solution should not be extremely dilute or have extreme pH, where the H⁺ or OH⁻ from water matters.
Strictly, activities should be used, but concentrations are a good approximation at low ionic strength.

Worked examples

pH of a buffer. 0.150 mol/L acetic acid and 0.250 mol/L sodium acetate: pH = 4.76 + log(0.250 ÷ 0.150) = 4.76 + 0.22 = 4.98.
Ratio needed. To make pH 4.40 from acetic acid: log(ratio) = 4.40 − 4.76 = −0.36, ratio = 0.44. So [A⁻] ÷ [HA] = 0.44.
Preparing a buffer from NaOH. 500 mL of a buffer at pH 5.00 with total 0.200 mol/L (0.100 mol in total) acetate species: ratio = 1.74, so A⁻ = 0.100 × 1.74 ÷ 2.74 = 0.0635 mol and HA = 0.0365 mol. Add 0.100 mol acetic acid and 0.0635 mol NaOH; the NaOH converts 0.0635 mol into acetate, leaving 0.0365 mol of acid.
Ammonia buffer. For NH₃/NH₄⁺ (pK_a of NH₄⁺ = 9.25), equal amounts give pH 9.25. With 0.30 mol NH₃ and 0.10 mol NH₄Cl: pH = 9.25 + log(3) = 9.73.
After adding strong acid. To 1.00 L of a buffer with 0.200 mol HA and 0.300 mol A⁻ (pK_a 4.76), 0.050 mol HCl is added: HA 0.250, A⁻ 0.250, pH = 4.76. Initially the pH was 4.76 + log(1.5) = 4.94.
Ionization of a drug. Aspirin (pK_a 3.5) in the stomach (pH 2.0): log([A⁻] ÷ [HA]) = 2.0 − 3.5 = −1.5, ratio 0.032, so 97% is un-ionized and passes through membranes easily. In the intestine (pH 6.5): log ratio = 3.0, ratio 1,000, so 99.9% is ionized.

How does it apply to blood?

The main buffer in blood is CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻, with an effective pK_a of 6.1 at 37 °C. The dissolved CO₂ concentration is 0.03 mmol/L per mmHg of CO₂ pressure. With [HCO₃⁻] = 24 mmol/L and pCO₂ = 40 mmHg (1.2 mmol/L dissolved CO₂): pH = 6.1 + log(24 ÷ 1.2) = 6.1 + 1.30 = 7.40. If breathing is impaired and pCO₂ doubles to 80 mmHg (2.4 mmol/L): pH = 6.1 + log(10) = 7.10, which is acidosis.

What about weak bases?

For a base B with conjugate acid BH⁺, use pH = pK_a(BH⁺) + log([B] ÷ [BH⁺]). Equivalent: pOH = pK_b + log([BH⁺] ÷ [B]).

Who formulated the equation?

Lawrence J. Henderson wrote an equation for [H⁺] in terms of the acid and salt concentrations in 1908. Karl Albert Hasselbalch expressed it in logarithmic form with pH in 1916 and applied it to blood. It is still a standard tool in physiology and pharmacology.

Which mistakes are common?

Mistake 1: Inverting the ratio

The base (A⁻) is in the numerator.

Mistake 2: Using it for a strong acid

It applies to weak acid/conjugate base pairs.

Mistake 3: Using initial concentrations after a reaction

After adding strong acid or base, recalculate the moles of HA and A⁻ first.

Mistake 4: Using it at extreme ratios

Outside 0.1 to 10, the approximations become poor.