Question

How many milliliters of a 0.339 M hydrochloric acid solution are required to neutralize 23.3 mL of a 0.668 M barium hydroxide solution?

Answer 1

С(HCl) = 0.339 M
V(Ba(OH)2) = 23.3 mL
C(Ba(OH)2) = 0.668 M
V(HCl) -?
2HCl + Ba(OH)2 => BaCl2 + 2H2O
ν(Ba(OH)2) = C(Ba(OH)2)* V(Ba(OH)2) = 0.668 M*0.0233 L = 0.016 mol
ν(HCl) = 2* ν(Ba(OH)2) = 2*0.016 mol = 0.032 mol
V(HCl) = ν(HCl)/ С(HCl) = 0.032 mol/ 0.339 M = 0.078 L = 78 mL