Titration: Back Titration

Question

33.7 mL of 1.54 M hydrochloric acid is added to 31.2 mL of sodium hydroxide, and the resulting solution is found to be acidic.

16.4 mL of 1.19 M barium hydroxide is required to reach neutrality.

What is the molarity of the original sodium hydroxide solution?

Answer 1

Frist reaction:

HCl + NaOH ==> NaCl + H2O

Since barium hydroxide (a base) was used in the back titration, that means HCl was in excess in the first reaction.

mmoles HCl present initially = 33.7 ml x 1.54 mmol/ml = 51.9 mmol HCl

Second reaction:

2HCl + Ba(OH)₂ ==> BaCl₂ + 2H₂O

mmoles Ba(OH)2 used = 16.4 ml x 1.19 mmol/ml = 19.52 mmoles

mmoles HCl present = 19.52 mmol Ba(OH)2 x 2 mmol HCl/mmol Ba(OH)2 = 39.0 mmol

This is the amount HCl left over after the first reaction with NaOH. So we can determine moles HCl

that reacted with the HCl as..

51.9 mmol - 39.0 mmol = 12.9 mmol HCl reacted in first reaction

mmoles NaOH reacted in first reaction = 12.9 since it is a 1:1 ratio with HCl

Molarity of original NaOH = 12.9 mmol/31.2 ml = 0.413 mmol/ml = 0.413 mol/L = 0.413 M