Calculate the pH of the titration mixture after 20.0 mL of base have been added.

Question

A 25.0 mL sample of a  0.100 M  solution of acetic acid is titrated with a  0.125 M  solution of NaOH. Calculate the pH of the titration mixture after 20.0 mL of base have been added. The K_a for acetic acid is 1.76×10^–5.

Answer 1

Initial moles HAc (acid acid) = 25.0 ml x 1 L / 1000 ml x 0.100 mol / L = 0.00250 mols HAc

Moles NaOH added = 20.0 ml x 1 L / 1000 ml x 0.125 mol / L = 0.00250 mols NaOH

HAc + NaOH ==> NaAc + H2O ... balanced equation

0.0025...0.0025........0.........................Initial

-0.0025...-0.0025......+0.0025.................Change

0................0...........0.0025..................Equilibrium (final)

Thus, we are at the equivalence point, where moles of base = moles acid. all of the HAc has been converted to the conjugate base (Ac^-). To find the pH, we need to consider the hydrolysis of the Ac^- anion.

Ac^- + H2O ==> HAc + OH^-

Kb = [HAc][OH^-] / [Ac^-] and Kb = 1x10^-14 / 1.76x10^-5 = 5.68x10^-10

5.68x10^-10 = (x)(x) / 0.0025 - x and assuming x is small relative to 0.0025, ignore it in the denominator

5.68x10^-10 = x² / 0.0025

x² = 1.42x10^-12

x = [OH^-] = 1.19x10^-6 M (note this is insignificant relative to 0.0025 so above assumption was valid)

pOH = -log 1.19x10^-6

pOH = 5.924

pH = 14 - pOH

pH = 8.076