Question

What volume of a 0.171 M barium hydroxide solution is required to neutralize 24.0 mL of a 0.225 M hydrobromic acid solution?

_________ mL barium hydroxide

Answer 1

 xmol    0.0054mol

Ba(OH)₂ + 2HBr = BaBr₂ +2H₂O

 1mol      2mol

 

n(HBr) = C(HBr) × V(HBr) = 0.225M × 0.024L = 0.0054mol.

n(Ba(OH)₂) = x = 0.0054 × 1 / 2 mol = 0.0027mol.

V(Ba(OH)₂) = n(Ba(OH)₂) / C(Ba(OH)₂) = 0.0027mol / 0.171M = 0.0158L = 15.8mL.