Prepare 250mL of 0.02 M Phosphate buffer in which [H2PO4]- = [HPO4]-
Calculate the volumes of 2 M phosphoric acid and 1 M NaOH required.
To a 250 ml volumetric flask containing about 100 ml of water, transfer the required volumes of 2 M phosphoric acid and 1 M NaOH, then fill to the mark with water and mix well.
Note that you should have written HPO4^=. You omitted a negative charge.
You want the buffer to be 0.02 M and you want 250 mL which is 0.02 x 0.250 L = 0.005 mols or 5 millimoles.
You want (H2PO4^-) + (HPO4^2-) = 5 millimoles in 250 mL = 0.02 M so you want (H2PO4^-) = 2.5 mmoles and (HPO4^2-) = 2.5 mmoles. Since you want equal amounts of each and it takes two steps you want to start with four times H3PO4. It looks like this.
...................H3PO4 + NaOH ==> NaH2PO4 + H2O
I..........4*2.5 = 10..........0..................0...........................
add...............................5.0................................................
Change.........-5.0..........-5.0................+5.0.........................
Equilibrium......5.0............0...................5.0..................
then the second step is
................NaH2PO4 + NaOH ==> Na2HPO4 + H2O
initial..........5.0..................0.................0................
add...................................2.5...........................
change.......-2.5.................-2.5...........+2.5
equilib..........2.5....................0...............2.5
so you want to start with 10 mmoles H3PO4 and add 7.5 mmoles NaOH which will leave you with 2.5 mmoles H2PO4^- and 2.5 mmoles HPO4^=.
You have 2 M H3PO4 = mmoles/mL or mL = mmoles/2 = 10.0 mmoles/2 = 5.00 mL of 2 M H3PO4. For NaOH you want to start with 7.5 mmoles of 1 M NaOH which is 7.5 mL of 1 M NaOH. So you plunk 5.00 mL of the 2 M H3PO4 and 7.5 mL of the 1 M NaOH into the 250 mL volumetric flask containing about 100 mL H2O, let cool, make to the mark with DI water, stopper, mix thoroughly, label, done. Check these calculations carefully.