[Borek]

Substitute into Henderson-Hasselbalch equation and solve.

[mrlucky0]

[H₂PO₄^-] = (2*0.1*x - 0.1*y) / (x+y)

Try the same approach to find out H₂PO₄^- concentration.

Where did this 2 come from?

Borek, I did follow your instructions and obtained the correct solution. But I'm a little unclear about the stoichiometry. Which two protons are you referring to that are being neutralized? From the reaction equation, isn't there only a transfer of 1 proton? Or, does the 2 come from the fact that H2PO4_- still has 2 protons left on it?

Can you clarify this part a bit more? Thanks, you've already been extremely helpful.

[Borek]

When you neutralize 1 proton you have H₂PO₄^- in the solution. You have to move further, hence the 2^nd.

[AWK]

In such a cases I teach my students to exchange protons (and Na+) stepwise:
H₃PO₄ + Na₃PO₄ = NaH₂PO₄ + Na₂HPO₄
Then, depending which reagent should be in excess
NaH₂PO₄ + Na₃PO₄ = 2Na₂HPO₄
or
H₃PO₄ + Na₂HPO₄ = 2NaH₂PO₄
The stoichiometry should be done in moles or in concentrations when all reagents are in the same volume.

[mrlucky0]

When you neutralize 1 proton you have H₂PO₄^- in the solution. You have to move further, hence the 2^nd.

Borek, let me see if I've got this straight. The HH equation calls for: log ([H₂PO₄^-] / [HPO₄^-2])

But we're really starting from H₃PO₄, and going to HPO₄^-2. Two protons are neutralized in this way.

If this reasoning was correct, I'm just glad I finally understood it. Unfortunately this concept was not very apparent to me. 

Edit:

In such a cases I teach my students to exchange protons (and Na+) stepwise:
H₃PO₄ + Na₃PO₄ = NaH₂PO₄ + Na₂HPO₄
Then, depending which reagent should be in excess
NaH₂PO₄ + Na₃PO₄ = 2Na₂HPO₄
or
H₃PO₄ + Na₂HPO₄ = 2NaH₂PO₄
The stoichiometry should be done in moles or in concentrations when all reagents are in the same volume.

Oh, that makes sense to me now.