It all comes down to what is a pKa. Until you have an intuition for pKa, you are doomed in later genchem and all of biochem.
As pKa is taught today, there are really only two things you need to know in order to answer 90% of pKa type questions.
When pH = pKa, then the acid is 50% dissociated.
Imagine you have pure water and a ziploc bag of crystallized dry acetic acid (like everything else in organic chemistry, it's a white powder). Wikipedia tells us that acetic acid has a pKa of 4.8. So you add a little acetic acid to the water. It gets more acidic. It's pH drops. You keep adding more, monitoring the pH. The pH goes from 7.0 to 5.9 to 4.8 to 3.7. Well, one thing that is true is that when the pH was 4.8, you know for sure that exactly half of the acetic acid molecules dissociated into H+ ions and acetate ions. The other half were just floating around as complete acetic acid molecules.
If you are building a buffer, you want the pH of the buffer to be close to the pKa of the acid.
The HH equation says "pH = pKa + log(A-/HA)". When you are building a buffer, you are making adjustments to the A- : HA ratio. If you start with exactly 50% of each, well, the log(1)=0 and your pH=pKa and your acid is 50% dissociated. Maybe you want a buffer with a pH of 5. Well, acetic acid would be a good choice. You would want just a teeny bit more of the conjugate base A- than the acid HA, which would adjust the pH slightly higher than the pKa of 4.8. Maybe you want a buffer with a pH of 2. You could still use acetic acid and have a MASSIVE amount of acetic acid and very little conjugate base like sodium acetate, but that would not be a very effective buffer because if you added more acid the pH would drop a lot. Buffers work best when they are close to 50% acid and 50% conjugate base.
So... knowing all this, let's look at your question now.
If the solution contains pure HPO4(2-), then its pH is average of pKa2 and pKa3
When the pH is pKa2, you know that the solution is half H2PO4(-) and half HPO4(2-).
When the pH is pKa3, you know that the solution is half HPO4(2-) and half PO4(3-).
Well, what do you suppose the solution looks like when you are half way between these two values? Yup, it is all HPO4(2-).
The pH continuum looks like this:
All H3PO4 | pKa1 | All H2PO4(-) | pKa2 | All HPO4(2-) | pKa3 | All PO4(3-)
