At 25 degree C, a student adds 0.400 g of solid phosphoric acid to a beaker containing 500.0ml of water. After the solid is completely dissolved the student adds 20.0ml of a 0.995M sodium hydroxide. What is the pOH of the solution in the beaker?

Answer 1

Interesting because at 25C H3PO4 is not solid. It freezes about 17 C.
mols H3PO4 = 0.4g/molar mass H3PO4 = estimated 0.004
mols NaOH added = M x L = estimated 0.02
You need to clean up the numbers before continuing.

......H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
Starting with 0.02 mols NaOH, how much Na3PO4 will be formed. That's 0.02/3 = 0.00667 mols Na3PO4.
Starting with 0.004 mols H3PO4, how much Na3PO4 will form? That's 0.004. In limiting regent problems the smaller number is ALWAYS the correct amount and the reagent producing that number is the limiting reagent. Therefore, the H3PO4 is the limiting reagent.
.....H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
I...0.004......0.........0........0
add.........0.02....................
C..-0.004..-0.012.......+0.004...+0.004
E.....0.....0.008.......0.0004...0.004

So you have a solution of about 0.008 mols NaOH in 520 mL which makes (OH^-) about 0.008/0.520 and convert to pOH. The PO4^3- formed will hydrolyze and contribute some OH^- but I doubt that will be significant.