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.
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.
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