Calculate the pH of the titration mixture after mL of base 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 10.0, 20.0, and 30.0 mL of base have been added. The K_a for acetic acid is 1.76×10^–5.

pH after 10.0 mL base added =

pH after 20.0 mL base added =

pH after 30.0 mL base added =

Answer 1

Acetic acid (HAc) is a weak acid. Sodium hydroxide is a strong base. Upon addition of NaOH to HAc, a buffer is formed until one reaches equivalence at which point all the HAc is converted to acetate (Ac^-) and H2O and the pH will be alkaline.

Initial moles HAc = 25.0 ml x 1 L / 1000 ml x 0.100 mol/L = 0.0025 mols HAc

Initial moles NaOH = 10.0 ml x 1 L / 1000 ml x 0.125 mol/L = 0.00125 mols NaOH (addition of 10 mls)

HAc + NaOH ==> NaAc + H2O

0.0025...0.00125.........0...............Initial

-0.00125...-0.00125..+0.00125....Change

0.00125.....0..............0.00125......Equilibrium

pH = pKa + log [NaAc]/[HAc] ... Henderson Hasselbalch equation

pKa = -log 1.76x10^-5 = 4.75

pH = 4.75 + log (0.00125/0.00125)

pH = 4.75 after addition of 10 mls of 0.125 M NaOH

After addition of 20 mls NaOH (20 ml x 1 L/1000 ml x 0.125 mol/L = 0.0025 mols NaOH)

HAc + NaOH ==> NaAc + H2O

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

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

0...........0.............0.0025......Equilibrium (this is at the equivalence point)

No longer a buffer as all HAc is converted to Ac^-

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

Kb for Ac^- = 1x10^-14 / 1.76x10^-5 = 5.68x10^-10

5.60x10^-10 = [HAc][OH^-] / [HAc] = (x)(x) / 0.0025

x² = 1.42x10^-12

x = [OH^-] = 1.19x10^-6

pOH = 5.92

pH = 8.08 (alkaline as predicted)

After addition of 30 ml NaOH (30 ml x 1 L/1000 ml x 0.125 mol/L = 0.00375 mols NaOH

HAc + NaOH ==> NaAc + H2O

0.0025...0.00375..........0............Initial

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

0..........0.00125..........0.0025.....Equilibrium

The [OH^-] from hydrolysis of 0.00125 mols Ac^- is small (1.19x10^-6 mols) relative to that from NaOH (0.00125 moles) so it can essentially be ignored.

moles NaOH = 0.00125 mols

final volume = 25 mls + 30 mls = 55 mls = 0.055 L

final [OH^-] = 0.00125 mols / 0.055 L = 0.0227 M

pOH = 1.64

pH = 12.36