Question

A 1.23 gram sample of an unknown monoprotic acid is dissolved in 25.0mL of water and titrated with a a 0.227 M aqueous barium hydroxidesolution. It is observed that after 12.1 milliliters of barium hydroxide have been added, the pH is 3.653 and that an additional 6.30 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 Ans.

Molarity of Ba(OH)₂ = 0.227M

Vol. of Ba(OH)₂ to reach equivalence point = 12.1mL + 6.30mL = 18.4mL

No. of mol of Ba(OH)₂ is

Since the acid neutralised is a monoprotic acid, 1 mol of a Ba(OH)₂ will neutralise 2 mol of the acid.

Therefore, the no. of mol of the monoprotic acid is twice the no. of mol of Ba(OH)₂ at the equivalence point.

The no. of mol of the monoprotic acid is

It is known that, the no. of mol is given by the formula

Therefore, the molecular weight of the monoprotic acid is 147g/mol.

2 Ans.

When 12.1mL of Ba(OH)₂ is added, the pH is 3.653

No. of mol of Ba(OH)₂ is

This much mol of the Ba(OH)₂ has no. of mol of OH^- given by

(since 1 mol of Ba(OH)₂ has 2 mol of OH^-)

The ICE table is

	HA	OH-	A-	H2O
I	8.3536 mmol	5.4934 mmol	0	-
C	-5.4934 mmol	-5.4934 mmol	+ 5.4934 mmol	-
E	2.8602 mmol	0	5.4934 mmol	-

pH formula is

Reaarange the equation

Therefore, the value of K_a for the acid is 4.27 x 10^-4