pH of formic acid and its salt

Question

An aqueous solution contains formic acid (pKs = 3.75) at 1.13 M and sodium formate (formic acid salt) at 2.41 M. What will be the pH if 2.35 grams of solid potassium hydroxide is added to a 200 ml portion of the solution?

Answer 1

Formic acid + sodium formate forms a buffer solution. Upon addition of potassium hydroxide (KOH), the KOH will react with the formic acid (HCOOH) to form formate (HCOO-) plus water.

Initial moles HCOOH = 1.13 mol/L x 0.200 L = 0.226 mol

Initial moles HCOO- = 2.41 mol/L x 0.200 L = 0.482 mol

Initial moles OH- = 2.35 g KOH x 1 mol KOH/56.1 g = 0.0419 mol

HCOOH + OH^- ==> HCOO^- + H2O

0.226.........0.0419.........0.482...........Initial

-0.0419......-0.0419......+0.0419........Change

0.1841.............0...........0.5239.........Equilibrium

From the Henderson Hasselbalch equation, we have

pH = pKa + log salt/acid

pH = 3.75 + log (0.5239/0.1841)

pH = 3.75 + 0.45

pH = 4.20