Balance the chemical equation given below, and determine the number of milliliters of 0.0300 M phosphoric acid required to neutralize 25.00 mL of 0.0150 M calcium hydroxide.

___ Ca (OH) 2 (aq) + ___ H3PO4 (aq) _ ___ Ca3 (PO4) 2 (s) + ___ H2O (l)

1) Balanced equation:

3 Ca (OH) 2 (aq) + 2 H3PO4 (aq) = 6 H2O (l) + Ca3 (PO4) 2 (s)

2) Calculations

Moles

H3 PO4:?

Ca[OH]2: 0.025 L * 0.0150 mol/L = 0.000375 mol

Ratios: 3 mol Ca[OH]2 / 2 mol H3PO4 = 0.000375 mol Ca[OH]2 / x mol H3PO4

=> x = 0.000375 * 2 / 3 = 0.00025 mol Ca[OH]2

Volume of Ca[OH]2 = # of moles / Molarity = 0.00025 mol / 0.0300 mol/L = 0.00833 L = 8.33 mL

Answer: 8.33 mL