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 the 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

Answer:

0.172 M

Explanation:

The reaction for the first titration is:

• HCl + NaOH → NaCl + H₂O

First we calculate how many HCl moles reacted, using the given concentration and volume:

• 19.6 mL * 0.189 M = 3.704 mmol HCl

As one HCl mol reacts with one NaOH mol, there are 3.704 NaOH mmoles in 25.0 mL of solution. With that in mind we determine the NaOH solution concentration:

• 3.704 mmol / 25.0 mL = 0.148 M

As for the second titration:

• H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

We determine how many NaOH moles reacted:

• 34.9 mL * 0.148 M = 5.165 mmol NaOH

Then we convert NaOH moles into H₃PO₄ moles, using the stoichiometric coefficients:

• 5.165 mmol NaOH * $\frac{1mmolH_3PO_4}{3mmolNaOH}$ = 1.722 mmol H₃PO₄

Finally we determine the H₃PO₄ solution concentration:

• 1.722 mmol / 10.0 mL = 0.172 M