How many milliliters of 2.75 M Ca(OH)2 is needed to completely neutralize 0.198 L of 2.03 M H3PO4. Make sure your answer has the correct number of significant figures.

Answer 1

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

mols H₃PO₄ = 0.198 L x 2.03 mol/L = 0.4019 mols H₃PO₄

mols Ca(OH)₂ needed = 0.4019 mols H₃PO₄ x 3 mols Ca(OH)₂ / 2 mol H₃PO₄ = 0.6029 mols Ca(OH)₂

Volume Ca(OH)₂ needed:

(x L)(2.75 mol/L) = 0.6029 mols

x = 0.219 L (3 sig. figs.)