Question

A 0.843 gram sample of an unknown monoprotic acid is dissolved in 40.0 mL of water and titrated with a a 0.496 M aqueous potassium hydroxidesolution. It is observed that after 5.16 milliliters of potassium hydroxide have been added, the pH is 9.092 and that an additional 9.84 mL of the potassium hydroxide solution is required to reach the equivalence point.

(1) What is the molecular weight of the acid? _______g/mol

(2) What is the value of K_a for the acid?

Answer 1

1)

Given data,

Molarity = 0.496 M

Total volume = 40 mL + 5.16 mL

= 45.16 mL

mol of base = Molarity x Total volume

= 0.496 x 45.16

= 22.39 mmol

mmol of acid = 22.39 mmol

Molecular weight = mass / mol of acid

= 0.843 / 22.39 x 10^-3

= 37.6 g / mol

2)

mmol of acid initially = 22.39 mmol

mmol of base initially = MV

= 0.496 * 5.16

= 2.55

mmol of acid left = 22.39 mmol - 2.55

= 19.84

mmol of conjugate formed = 0 + 2.55

= 2.55

pH =  pKa + log(2.55 / 19.84 )

pKa = 9.092 - 0.892

pKa = 8.2