Henderson–Hasselbalch Equation

pH = pKa + log([A−]/[HA]) predicts the pH of a buffer from the ratio of conjugate base to weak acid.

pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)

pH = pKa + log([A-] / [HA])

What it means

The Henderson–Hasselbalch equation says a buffer's pH is anchored at the acid's pKa and nudged by the base-to-acid ratio. When base and acid are equal, the log term is zero and pH = pKa — the buffer's center. More base than acid pushes the pH above pKa; more acid pushes it below. Because the ratio sits inside a logarithm, changing it tenfold shifts the pH by exactly one unit.

Variables

Symbol	Name	Unit	Meaning
$pH$	Solution pH	unitless	The pH of the buffer solution being calculated.
$pKa$	Acid's pKa	unitless	The negative log of the weak acid's Ka; the pH the buffer holds when base and acid are equal.
$[A⁻]$	Conjugate base concentration	mol/L (M)	The molar concentration of the conjugate base component of the buffer.
$[HA]$	Weak acid concentration	mol/L (M)	The molar concentration of the weak acid component of the buffer.

Assumptions

The buffer contains a weak acid and its conjugate base (or vice versa).
The solution is dilute enough that concentration approximates activity.
The acid-base equilibrium is established at constant temperature.
The equation is most reliable when the base-to-acid ratio is between 0.1 and 10, where buffer capacity is good.

Derivation

Start from Ka = [H⁺][A⁻]/[HA] and solve for [H⁺]: [H⁺] = Ka × [HA]/[A⁻]. Take the negative base-10 logarithm of both sides: −log[H⁺] = −log Ka − log([HA]/[A⁻]), which is pH = pKa + log([A⁻]/[HA]). The ratio flips sign inside the log to give the final form.

Worked example

Problem

An acetate buffer contains 0.200 M CH3COO⁻ (conjugate base) and 0.100 M CH3COOH (weak acid, Ka = 1.8 × 10⁻⁵). Calculate the buffer pH.

Given:

[CH3COO⁻] = 0.200 M
[CH3COOH] = 0.100 M
Ka = 1.8 × 10⁻⁵ for acetic acid

First find pKa = −log Ka = −log(1.8 × 10⁻⁵).
log(1.8 × 10⁻⁵) = log(1.8) + log(10⁻⁵) = 0.255 − 5 = −4.745, so pKa = 4.74.
Apply Henderson–Hasselbalch: pH = 4.74 + log(0.200 / 0.100).
The ratio is 0.200 / 0.100 = 2.00, and log(2.00) = 0.301.
pH = 4.74 + 0.301 = 5.04.

pH = 5.04

The buffer is base-rich, so the pH sits above pKa, and the 2:1 ratio lies inside the effective buffering range — this mixture resists pH change well.