50 mL of 0.2 M KOH is added to 150 mL of 0.1 M acetic acid. What is the pH change of the acetic acid solution caused by the addition of KOH? (Ka of acetic acid is 1.8x10-5)

Answer 1

When you add KOH (a base) to acetic acid (a weak acid), you end up making a buffer. We can write the equation for this reaction as follows:

KOH + CH₃COOH ==> CH₃COOK + H2O

First, we will calculate the pH of the acetic acid alone, before addition of KOH.

Let HAc represent the acetic acid.

HAc ==> H⁺ + Ac^- Ka = 1.8x10^-5

Ka = [H⁺][Ac] / [HAc]

1.8x10^-5 = (x)(x) / 0.1 - x and assume x is small compared to 0.1 M and ignore it

1.8x10^-5 = x² / 0.1

x = [H⁺] = 1.34x10^-3 M (this is about 1.3% of 0.1 M so above assumption was valid)

pH = -log [H⁺] = -log 1.34x10^-3

pH = 2.87

Next, we will calculate the pH after adding the KOH:

moles HAc present initially = 150 ml x 1 L / 1000 ml x 0.1 mol / L = 0.015 mols HAc

moles KOH added = 50 ml x 1 L / 1000 ml x 0.2 mol / L = 0.01 mols KOH

moles HAc after reaction = 0.015 mol - 0.01 mol = 0.005 mols HAc

moles NaAc formed = 0.01 mols NaAc

This may better be seen using an ICE table:

KOH + CH₃COOH ==> CH₃COOK + H2O

0.01......0.015......................0................0........Initial

-0.01....-0.01......................+0.01....................Change

0...........0.005......................0.01.....................Equilibrium

The pH of this buffer (HAc + NaAc) can be determined using the Henderson Hasselbalch equation:

pH = pKa + log [NaAc] / [HAc] and pKa = -log Ka = -log 1.8x10^-5 = 4.74

pH = 4.74 + log (0.01 / 0.005) = 4.74 + log 2

pH = 4.74 + 0.30

pH = 5.04

Thus, the CHANGE in pH = 5.04 - 2.87 = 2.17 pH units