Henderson-Hasselbalch Equation and Example

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):

_apH = 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 HC₂H₃O₂ and 0.50 M C₂H₃O₂^- that has an acid dissociation constant for HC₂H₃O₂ 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 ([C₂H₃O₂^-] / [HC₂H₃O₂])

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