Question

1. When a 19.6 mL sample of a 0.498 M aqueous acetic acid solution is titrated with a 0.343 M aqueous potassium hydroxide solution, what is the pH after 42.7 mL of potassium hydroxide have been added?

pH=

2.What is the pH at the equivalence point in the titration of a 22.2 mL sample of a 0.413 M aqueous acetic acid solution with a 0.500 M aqueous potassium hydroxide solution?

pH=

Answer 1

1)

Millimoles of acetic acid = volume of acetic acid x molarity of acetic acid

= 19.6 mL x 0.498 M

= 9.7608 mmol

Millimoles of KOH added = Volume of KOH x molarity of KOH

= 42.7 mL x 0.343 M

= 14.6461 mmol

Reaction:

CH₃COOH + KOH ------> CH₃COOK + H₂O

After reaction:

Since there are excess moles of KOH, the pH of the solution after the reaction is due to strong base (KOH).

Millimoles of KOH remains = 14.6461 mmol - 9.7608 mmol = 4.8853 mmol

Concentration of OH^- = millimoles of KOH/total volume in mL = 4.8853 mmol/(19.6 mL + 42.7 mL) = 0.0784 mmol/mL = 0.0784 M

pOH = -log[OH^-] = - log (0.0784) = 1.11

pH = pKw - pOH = 14.00 - 1.11 = 12.89