Question

A solution contains 10. mmol of H3PO4 and 5.0 mmol of NaH2PO4. How many milliliters of 0.10 M NaOH must be added to reach the second equivalence point of the titration of the H3PO4 with NaOH

Answer 1

Explanation:

Reaction equations for the given species is as follows.

   $H_{3}PO_{4} + 2NaOH \righleftharpoons Na_{2}HPO_{4} + 2H_{2}O$

   $NaH_{2}PO_{4} + NaOH \rightleftharpoons Na_{2}HPO_{4} + H_{2}O$

At the first equivalence point we need 2 × 10 mmol NaOH.

At the second equivalence point we need 5 mmol of NaOH.

Hence, total moles of NaOH required is as follows.

               (20 + 5) mmol = 25 mmol

We assume that volume of NaOH required is V.

      $25 mmol NaOH \times \frac{10^{-3} mol NaOH}{1 mmol NaOH} \times \frac{1000 ml V}{0.10 \text{mol NaOH}}$

      = 250 ml V

Thus, we can conclude that 250 ml of 0.10 M NaOH must be added to reach second equivalence point of the titration of the $H_{3}PO_{4}$ and NaOH.