What volume in milliliters of 0.0160 M Ca(OH)₂ is required to neutralize 75.0 mL of 0.0300 M HCl?

Answer 1

Ca(OH)₂ + 2HCl ==> 2H₂O + CaCl₂ ... balanced equation

moles HCl present = 75.0 ml x 1 L / 1000 ml x 0.0300 mol/L = 0.00225 mols HCl

moles Ca(OH)₂ needed = 0.00225 mols HCl x 1 mol Ca(OH)₂ / 2 mols HCl = 0.001125 mols Ca(OH)₂

volume (in mls) of 0.016 M Ca(OH)₂ = 0.001125 mols Ca(OH)₂ x 1 L / 0.016 mol x 1000 ml/L = 70.3 mls