Question

A 1.75 gram sample of an unknown monoprotic acid is dissolved in 40.0 mL of water and titrated with a a 0.477 M aqueous barium hydroxide solution. It is observed that after 9.85 milliliters of barium hydroxide have been added, the pH is 3.603 and that an additional 5.05 mL of the barium 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)

Answer

123.12g/mol

Explanation

The reaction between monoprotic acid and barium hydroxide is as follows

2HA + Ba(OH)2 -------> B(A)2 + 2H2O

stoichiometrically, 2moles of HA react with 1mole of Ba(OH)2 ,

Equivalence point = 14.90ml

number of moles of Ba(OH)2 consumed = (0.477mol/1000ml) ×14.90ml = 0.007107mol

moles of HA = 2× 0.007107mol = 0.014214mol

molar mass = mass/number of moles

molar mass of the monoprotic acid = 1.75g/0.014214mol = 123.12g/mol

2)

Answer

Ka = 4.86×10^-4

Explanation

Number of moles of Ba(OH)2 added = (0.477mol/1000ml)×9.85mol =0.004698mol

Number of moles of HA reacted = 2× 0.004698mol = 0.009396mol

number of moles of A^- formed = 0.009396mol

remaining moles of HA = 0.014214mol - 0.009396mol = 0.004818

Total volume =.49.85ml

[A^-] =( 0.009396mol/49.85ml)×1000ml = 0.1885M

[HA] = (0.004818mol/49.85ml)×1000ml = 0.09665M

Henderson - Hasselbalch equation

pH = pKa + log([A^-]/[HA])

3.603 = pKa + log( 0.1885M/0.09665M)

pKa = 3.603 - 0.290 = 3.313

pKa = -logKa

-logKa = 3.313

Ka = 4.86×10^-4