Question

An analytical chemist is titrating 60.5mL of a 0.8700M solution of benzoic acid HC6H5CO2 with a 0.3600M solution of KOH . The pKa of benzoic acid is 4.20 . Calculate the pH of the acid solution after the chemist has added 172.mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places.

Answer 1

SOLUTION:

KOH will react with benzoic acid to form benzoate, as a result the solution will behave as buffer because a weak acid and its conjugated base are present in the solution:

Moles of benzoic acid = Molarity X volume of solution = 0.8700M X 60.5mL = 52.635 mmol = 0.052635 Moles

Moles of KKOH added = 0.3600 X 172 = 61.92 mmoles = 0.06192 Moles

Out of 61.92 mmoles of KOH 52.635 will react with benzoic acid.

Hence moles of KOH that remain unracted = 61.92 mmoles - 52.635 mmol = 9.285 mmoles = 0.009285 Moles

Molarity of KOH = Number of moles / Volume of solution in liters

Total Volume = 172mL + 60.5 = 232.5mL = 0.2325L

Molarity of KOH = 0.009285 Moles / 0.2325L = 0.0395M

KOH dissociates completely to form OH-

pOH = -log[0.0395] = 1.4

pH + pOH = 14

pH = 14 - 1.4 = 12.6