H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(l)

Question

A solution of 0.316 M KOH is used to neutralize 18.0 mL H3PO4 solution.

If 30.3 mL KOH solution is required to reach the endpoint, what is the molarity of the H3PO4 solution?

Answer 1

Neutralization of an acid (H₃PO₄) and a base (KOH)

H₃PO₄(aq) + 3KOH(aq) ==> K₃PO₄(aq) + 3H₂O(l) ... note mol ratio of base: acid

moles KOH used = 30.3 ml x 1L/1000 ml x 0.316 mol/L = 0.0095 75 moles KOH

moles H₃PO₄ present = 0.009575 mol KOH x 1 mol H₃PO₄/3 mol KOH = 0.003192 moles H₃PO₄

Molarity of H₃PO₄ = 0.003192 moles/0.0180 L = 0.1773 M = 0.177 M (to 3 significant figures)