Find the pH and pOH of an acetic acid solution

Question

Acetic (ethanoic) acid is the most common weak acid used in industry.  Determine the pH and pOH of an acetic acid solution prepared by dissolving 60.0 kg of pure, liquid acetic acid, to make 1.25 kL of solution.  The percent reaction with water at this concentration is 0.48%. You must show ALL your work to get full marks.

Answer 1

pH = -log [H⁺]

pOH = -log [OH^-]

CH3COOH ==> CH3COO^- + H⁺

Let CH3COOH be represented by HAc

molar mass HAc (acetic acid) = 60.0 g/mol

60.0 kg HAc x 1000 g / kg x 1 mol / 60.0 g = 1000 mols HAc

Molarity = moles/liter and 1.25 kL = 1250 L

Molarity = 1000 moles / 1250 L = 0.8 M

Percent ionization/dissociation = 0.48%

0.48% x 0.8M = 0.0038 M H⁺ and 0.0038 M Ac^-

pH = -log H+] = -log 0.0038

pH = 2.42

pOH = 14 - pH

pOH = 11.6