Question

A 1.35 L buffer solution consists of 0.173 M butanoic acid and 0.309 M sodium butanoate. Calculate the pH of the solution following the addition of 0.076 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.

Answer 1

Ka of butanoic acid = 1.52*10^-5
pka = -log Ka
= -log (1.52*10^-5)
= 4.82

mol of butanoic acid present initially = M*V = 0.173 M * 1.35 L = 0.234 mol
mol of sodium butanoate present initially = 0.309 * 1.35 =0.417 mol

mol of NaOH added = 0.076 mol

0.076 mol of butanoic acid and sodium butanoate will react to form extra 0.076 mol of sodium butanoate

after reaction,
mol of sodium butanoate= 0.417 + 0.076 = 0.493 mol
mol of butanic acid remaning = 0.234 - 0.076 mol = 0.158 mmol

total volume = 1.35 L
[sodium butanoate] = 0.493/1.35 = 0.365 M
[butanoic acid] = 0.158/1.35 = 0.117 M

use:
pH = pKa + log {[sodium butanoate]/[butanoic acid]}
= 4.82 + log (0.365/0.117)
= 5.31
Answer: 5.31