An aqueous solution of hydrobromic acid is standardized by titration with a 0.173 M solution of potassium hydroxide.

Question

An aqueous solution of hydrobromic acid is standardized by titration with a 0.173 M solution of potassium hydroxide.

If 18.9 mL of base are required to neutralize 29.7 mL of the acid, what is the molarity of the hydrobromic acid solution?

Answer 1

HBr(aq) + KOH(aq) ==> H2O(l) + KBr(aq) ... balanced equation

mols KOH used = 18.9 ml x 1 L / 1000 ml x 0.173 mol / L = 0.0032697 mols KOH used

mols HBr present = 0.0032697 mols KOH x 1 mol HBr / mol KOH = 0.0032697 mols HBr

Molarity of HBr = mols / L = 0.0032697 mols / 29.7 mls x 1000 ml / L = 0.110 mols/L = 0.110 M HBr (3 s.f.)