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Question

A 1.42 L buffer solution consists of 0.168 M butanoic acid and 0.302 M sodium butanoate. Calculate the pH of the solution following the addition of 0.061 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The 𝐾a of butanoic acid is 1.52×10^−5.

pH=?

Answer 1

moles HA = 1.42 L x 0.168 mol / L = 0.2386 mols

moles A- = 1.42 L x 0.302 mol / L = 0.4288 mols

mols OH- added = 0.061.

HA + OH- ==> A- + H2O

0.2386...0.061......0.4288............Initial

-0.061...-0.061......+0.061...........Change

0.1776.......0...........0.4898.........Equilibrium

Henderson Hasselbalch equation

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

pKa = -log Ka = -log 1.52x10^-5 = 4.82

pH = 4.82 + log (0.4898 / 0.1776) = 4.82 + 0.44

pH = 5.26