How much 2.00 M H3PO4, in mL, would you need to add to 50.00 mL of 3.50 M Ca(OH)2 in order to neutralize the solution?

Answer 1

Looking at the balanced equation:

2H₃PO₄ + 3Ca(OH)₂ ==> Ca₃(PO₄)₂ + 6H₂O ... balanced equation

Next, find moles of Ca(OH)₂ present:

50.00 ml x 1 L / 1000 mls x 3.50 mol/L = 0.175 mols

Next, find moles of H₃PO₄ that will react with 0.175 mols Ca(OH)₂:

0.175 mol Ca(OH)₂ x 2 mols H₃PO₄ / 3 mols Ca(OH)₂ = 0.1167 mols

Finally, find the volume of 2.00 M H₃PO₄ needed to supply 0.1167 mols of H₃PO₄:

0.1167 mols H₃PO₄ x 1 L / 2.00 mols = 0.05835 L = 58.35 mls H₃PO₄