Question

A 1.22 gram sample of an unknown monoprotic acid is dissolved in 25.0 mL of water and titrated with a a 0.299 M aqueous potassium hydroxide solution. It is observed that after 10.4 milliliters of potassium hydroxide have been added, the pH is 3.039 and that an additional 19.4 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 Ka for the acid?

Answer 1

MM = mass /mol of acid = 1.22 g / mol of acid

find mol of acid via:

mol of base = mol of acid

so

mol of base = M*V = (0.299) * (10.4 +19.4) = 8.9102 mmol of base

mmol of acid = 8.9102

MM = mass /mol of acid = 1.22 g / mol of acid

MM = mass /mol of acid = 1.22 / (8.9102 *10^-3) = 136.92 g/mol

b)

find Ka:

via half equivalence point

pH = pKa + log(A-/HA)

mmol of acid initially = 8.9102

mmol of base initially = MV = 0.299*10.4 = 3.1096

mmol of acid left = 8.9102 -3.1096 = 5.8006 mmol of acid

mmol of conjugate formed = 0 + 3.1096 = 3.1096

so

3.039 = pKa + log(3.1096/5.8006 )

pKa = 3.039 - log(3.1096/5.8006 )

pKa = 3.3097

Ka = 10^-pKA = 10^-3.3097 = 49011*10^-4