You can calculate the pH of a buffer solution or the concentration of the acid and base using the Henderson-Hasselbalch equation. Here's a look at the Henderson-Hasselbalch equation and a worked example that explains how to apply the equation.

## Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates pH, pKa, and molar concentration (concentration in units of moles per liter):

_{a}pH = pK + log ([A^{-}]/[HA])

[A^{-}] = molar concentration of a conjugate base

[HA] = molar concentration of an undissociated weak acid (M)

The equation can be rewritten to solve for pOH:

**pOH = pK _{b} + log ([HB^{+}]/[ B ])**

[HB^{+}] = molar concentration of the conjugate base (M)

[ B ] = molar concentration of a weak base (M)

## Example Problem Applying the Henderson-Hasselbalch Equation

Calculate the pH of a buffer solution made from 0.20 M HC_{2}H_{3}O_{2} and 0.50 M C_{2}H_{3}O_{2}^{-} that has an acid dissociation constant for HC_{2}H_{3}O_{2} of 1.8 x 10^{-5}.

This is a straightforward example because all of the terms are given. Solve this problem by plugging the values into the Henderson-Hasselbalch equation for a weak acid and its conjugate base.

pH = pK_{a} + log ([A^{-}]/[HA])

pH = pK_{a} + log ([C_{2}H_{3}O_{2}^{-}] / [HC_{2}H_{3}O_{2}])

pH = -log (1.8 x 10^{-5}) + log (0.50 M / 0.20 M)

pH = -log (1.8 x 10^{-5}) + log (2.5)

pH = 4.7 + 0.40

pH = 5.1