Question

A 25.0 mL NaOH solution of unknown concentration was titrated with a 0.189 M HCl solution. 19.6 mL HCl was required to reach equivalence point. In a separate titration, a 10.0 mL H₃PO₄ solution was titrated with the same NaOH solution. This time, 34.9 mL NaOH was required to reach the equivalence point. What is the concentration of the H₃PO₄ solution?

Answer 1

For the first titration:

Volume of HCl (V₁) = 19.6 mL

Concentration of HCl (C₁) = 0.189 M

Volume of NaOH (V₂) = 25.0 mL

Let us say that the concentration of NaOH = C₂

Equating the moles of acid and base to get,

V₁C₁ = V₂C₂

or, 19.6 mL x 0.189 M = 25.0 mL x C₂

or, C₂ = 0.148 M

Therefore, the concentration of NaOH = 0.148 M

For the second titration:

Volume of NaOH (V₁') = 34.9 mL

Concentration of NaOH (C₁') = 0.148 M

Volume of H₃PO₄ (V₂') = 10.0 mL

Let us say that the concentration of H₃PO₄ = C₂'

Equating the moles of acid and base to get,

V₁'C₁'= V₂'C₂'

or, 34.9 mL x 0.148 M = 10.0 mL x C₂'

or, C₂' = 0.517 M

Hence, the concentration of the H₃PO₄ solution = 0.517 M