Question

A 12.1 g sample of an aqueous solution of hydrochloric acid contains an unknown amount of the acid.
If 14.3 mL of 0.856 M barium hydroxide are required to neutralize the hydrochloric acid, what is the percent by mass of hydrochloric acid in the mixture?

Answer 1

a)

m = 12.1 g sample

x mol of HCl

V = 14.3 mL of Ba(OH)2 solution

M = 0.865 M of Ba(OH)2

if neutralized

Ba(OH)2 + HCl = H2O + BaCl2

balance

Ba(OH)2 + 2HCl = 2H2O + BaCl2

then,

mol of base = 2 mol of acid

mol of Ba(OH)2 = MV = 0.856*(14.3*10^-3) = 0.0122408 mol of Ba(OH)2

mol of HCl = 2* mol of Ba(OH)2 = 2* 0.0122408 = 0.0244816 mol of HCl

now...

mass of HCl = mol*MW = 0.0244816 *36.4609 = 0.89262 g of HCl

% mass of HCl = mass of HCl / Total mass * 100% = 0.89262 / 12.1 *100 = 7.3770% is HCl