A 4.0520 g sample of hydrochloric acid, sp gr 1.18, required 44.15 ml of 0.9035 N sodium hydroxide in a titration.

Question

Calculate the HCl content in %

Answer 1

Something seem amiss with this question. If you really have 4.0520 g of HCl, that equates to

4.0520 g x 1 mol HCl / 36.46 g = 0.1111 moles HCl. This should require 0.1111 moles of NaOH to neutralize since the reaction is HCl + NaOH ==> NaCl + H2O

However, moles NaOH computed does not equal 0.1111 moles.

moles NaOH = 44.15 mls x 1 L /1000 mls x 0.9035 mol/L = 0.03989 mols NaOH

Something is wrong with the problem. By the way, the specific gravity (1.18) does not enter into the problem since we already know the mass of HCl. Unless you want the %HCl by volume and not by mass.

Even then, there is something wrong with the problem as written.