How many milliliters of 0.318 M Ba(OH)2 solution are needed to completely neutralize 39.0 mL of 0.181 M HCl solution? The reaction is Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2H2O(l)

Answer 1

Ba(OH)₂ + 2HCl ==> BaCl₂ + 2H₂O ... balanced equation

moles HCl present = 39.0 ml x 1 L / 1000 ml x 0.181 mols/L = 0.007059 moles HCl'

moles Ba(OH)₂ needed = 0.007059 mols HCl x 1 mol Ba(OH)₂ / 2 mols HCl = 0.003530 moles Ba(OH)₂ needed

volume (mls) Ba(OH)₂ needed = 0.003530 mols Ba(OH)₂ x 1 L / 0.318 mols = L x 1000 ml/L = 11.1mls needed