In a titration, 15.94 mL of 0.512 M KOH solution is required to reach the endpoint of a 50.00-mL sample of a phosphoric acid, H3PO4, solution. Calculate the molarity of the H3PO4 solution.

Answer 1

Molarity is moles per liter of solution. We have the liters of H₃PO₄ given as 50.00 ml x 1 L/1000 ml = 0.0500 L. So we now need to find how many moles of H₃PO₄ are present. Looking at the balanced equation for the titration, we have...

3KOH + H₃PO₄ ==> 3H₂O + K₃PO₄

moles KOH used in titration = 15.94 ml x 1 L/1000 ml x 0.512 mol/L = 0.008161 moles KOH

moles H₃PO₄ neutralized = 0.008161 mol KOH x 1 mol H₃PO₄/3 mol KOH = 0.002720 moles H₃PO₄

Molarity of H₃PO₄ = 0.002720 mol/0.0500 L = 0.0544 M (to 3 significant figures based on 0.512 M having 3 significant figures)