Question

A 35.00 mL sample of an unknown H3PO4solution is titrated with a 0.130 M NaOH solution. The equivalence point is reached when 25.83 mLof NaOH solution is added.

What is the concentration of the unknown H3PO4 solution? The neutralization reaction is

H3PO4(aq)+3NaOH(aq)→3H2O(l)+Na3PO4(aq)

Answer 1

Ans. Balanced reaction: H₃PO₄(aq) + 3 NaOH (aq) ------> Na₃PO₄(aq) + 3 H₂O

At equivalence point, the total number of moles of H⁺ from acid is equal to total number of OH^- from base.

That is-           x(M1V1), acid = y(M2V2), base

            Where, x = moles of H⁺ produced per mol acid = 3 for H₃PO₄

                        y = moles of OH^- produced per mol base = 1 for NaOH

                        V and M are volume and molarity of respective solution.

Putting the values in above expression-

                        3 (M1 x 35.0 mL) = 1 x (0.130 M x 25.83 mL)

                        Or, M1 = (0.130 M x 25.83 mL) / (3 x 35.0 mL)

                        Therefore, M1 = 0.03198 M

Therefore, molarity of H₃PO₄ = 0.03198 M

# Method 2:

Balanced reaction: H₃PO₄(aq) + 3 NaOH (aq) ------> Na₃PO₄(aq) + 3 H₂O

# Moles of NaOH consumed = Molarity x Volume of solution in liters

                                                = 0.130 M x 0.02583 L

                                                = 0.0033579 mol

# According to the stoichiometry of balanced reaction, 1 mol H₃PO₄ neutralizes 3 mol NaOH.

So,

Required moles of H₃PO₄ = (1/3) x Moles of NaOH

                                                = (1/3) x 0.0033579 mol

                                                = 0.0011193 mol

# Now,

            Molarity of H₃PO₄ = Moles of H₃PO₄ / Volume of solution in liters

                                                = 0.0011193 mol / 0.0350 L

                                                = 0.03198 M