Question

Question 6: Postlecture 11

A 1.45 L buffer solution consists of 0.189 M butanoic acid and 0.302 M sodium butanoate. Calculate the pH of the solution following the addition of 0.066 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The ?a of butanoic acid is 1.52×10−5 .

pH=

Answer 1

mol of NaOH added = 0.066 mol

C3H7COOH will react with OH- to form C3H7COO-

Before Reaction:
mol of C3H7COO- = 0.302 M *1.45 L
mol of C3H7COO- = 0.4379 mol

mol of C3H7COOH = 0.189 M *1.45 L
mol of C3H7COOH = 0.2741 mol

after reaction,
mol of C3H7COO- = mol present initially + mol added
mol of C3H7COO- = (0.4379 + 0.066) mol
mol of C3H7COO- = 0.5039 mol

mol of C3H7COOH = mol present initially - mol added
mol of C3H7COOH = (0.2741 - 0.066) mol
mol of C3H7COOH = 0.2081 mol


Ka = 1.52*10^-5

pKa = - log (Ka)
= - log(1.52*10^-5)
= 4.818

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.818+ log {0.5039/0.2081}
= 5.202

Answer: 5.20