what is the molarity of an H3PO4 solution if 135 mL is completely reacted by 674 mL of 0.200 M Ca(OH)2 solution?

Answer 1

First, write the correctly balanced equation for the reaction taking place:

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

Next, find the moles of Ca(OH)2 used:

moles Ca(OH)₂ = 674 ml x 1 L / 1000 ml x 0.200 mol / L = 0.1348 mols Ca(OH)₂

Next, find moles of H3PO4 present:

moles H₃PO₄ = 0.1348 mols Ca(OH)₂ x 2 molH₃PO₄ / 3 mol Ca(OH)₂ = 0.08987 mols H₃PO₄

Finally, calculate the molarity of H₃PO₄:

molarity = moles / liters

molarity = 0.08987 mols / 135 mls x 1000 mls / L = 0.666 mols / L (3 sig. figs.)