Question

PART A: A buffer is created by mixing 0.10 mol of acetic acid and 0.13 mol of sodium acetate. What is the pH of this solution? PART B: If the solution from part A has 0.02 mol of KOH added to it, what is the new pH?

Answer 1

The Ka of acetic acid = 1.8 x 10^-5

The Henderson Hasselbach equation =

pH = -log pKa + log [base/acid]

[ acetic acid] = 0. 1

[sodium acetate] = 0.13mol

pH = -logPka + log[0.13/0. 1] =

= -log(1.8 x 10^-5) +0.1139 = 4.746 + 0.1139 = 4.858

b.

The 0.02mol KOH base will convert 0.02mol of acetic acid to the conjugate base acetate

Therefore the concentration of sodium acetate = 0.13 mol + 0.02mol = 0.15 mol base

During this process it used 0.02mol acid leaving behind 0.1mol - 0.02mol = 0.08mol acid

As per the Henderson-Hasselbalch equation

pH = -log(1.8X10^-5) + log(0.15mol / 0.08mol) =

= 4.744 + 0.273 = 5.02