In: Chemistry
Calculate the volume of 0.20 mol/L KH2PO4 and the volume of 0.10 mol/L NaOH required to prepare 50.0 mL of buffer with a formal (or total) concentration of H2PO4 of 0.020 mol/L and a pH = 6.3. Repeat this calculation for buffers (of the same volume and formal concentration) with pH = 6.5, 6.9, 7.5, 7.9, and 8.2.
H2PO4^- + NaOH ----------------------------> HPO4^-2 + H2O
0.2 0.1 0 0
0.2-0.1 0 0.1 0.1
so NaOH molarity = HPO4^- molairty
millimoles of total buffer = [H2PO4-] + [HPO4-2] = 0.02 x 50 = 1
total volume = 50
H2PO4 volume = x
HPO4^-2 = 50 -x
H2PO4- millimoles = 0.2 x
HPO4-2 millimoles = 0.1 (50 -x)
pH = pKa2 + log [salt /acid]
6.3 = 7.20 + log (0.1 (50 -x) / 0.2 x)
0.126 = 0.1 (50 -x) / 0.2 x
0.025 x = 5 - 0.1 x
x = 39.94 ml
50 -x = 10.06 ml
KH2PO4 volume = x = 39.94 ml
NaOH volume = 50 -x = 10.06 ml