How many mL of 2.0 M Ca(OH) 2 are required to completely neutralize 100 mL of 4.0 M H3PO4

Answer 1

Always write the balanced equation:

3Ca(OH)₂ + 2H₃PO₄ ==> Ca₃(PO₄)₂ + 3H₂O

Note the mol ratio of 3 : 2 for Ca(OH)₂ : H₃PO₄

moles H₃PO₄ present = 4.0 mol/L x 0.100 L = 0.40 moles

moles Ca(OH)₂ needed = 0.40 mol H₃PO₄ x 3 mol Ca(OH)₂ / 2 mol H₃PO₄ = 0.6 mol Ca(OH)₂ needed

volume Ca(OH)₂ needed: (x L)(2.0 mol/L) = 0.6 moles

x = 0.3 L = 300 mls of 2.0 Ca(OH)₂ needed