ChemixaLEARN CHEMISTRY

pH of an Acetic Acid/Acetate Buffer

Medium
Problem

A buffer is prepared containing 0.10 M acetic acid (CH3COOH) and 0.10 M sodium acetate (CH3COO-). Calculate the pH of this buffer at 25 C (pKa of acetic acid = 4.74).

Given

  • [CH3COOH] (acid) = 0.10 M

  • [CH3COO-] (conjugate base) = 0.10 M

  • pKa(acetic acid) = 4.74 at 25 C

Find

The pH of the buffer solution.

Principle

A buffer contains a weak acid and its conjugate base together. Starting from Ka = [H+][A-]/[HA] and taking the negative logarithm gives the Henderson-Hasselbalch equation: pH = pKa + log([base]/[acid]). When the two concentrations are equal, the log term is zero and pH = pKa, the point of maximum buffering capacity.

Formula

Solution

  1. Identify acid and conjugate base: HA = CH3COOH at 0.10 M, A- = CH3COO- at 0.10 M.

  2. Write the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]).

  3. Substitute: pH = 4.74 + log(0.10/0.10) = 4.74 + log(1).

  4. Since log(1) = 0, the correction term vanishes and pH = 4.74.

pH = 4.74

Unit check

The ratio [base]/[acid] has units M/M, which cancel, so the logarithm acts on a dimensionless number and pH is dimensionless, as required.

What this means physically

When acid and conjugate base are present in equal amounts, a buffer's pH equals the acid's pKa — 4.74 here. This buffer is most effective near pH 4.74: added H+ is consumed by the acetate base and added OH- by the acetic acid, keeping the pH nearly steady.

Common mistake

Students frequently invert the ratio, writing log([acid]/[base]) instead of log([base]/[acid]), which flips the sign of the correction. Another error is using Ka instead of pKa in the equation.