The dominant form of the phosphoric acid is #"HPO"_4^(2-)# and its concentration is #"0.639 mmol/L"#.
So, start with your three balcanced chemical equations
#H_3PO_4 + H_2O rightleftharpoons H_2PO_4^(-) + H_3^(+)O#, #pKa_1 = 2.12#
#H_2PO_4^(-) + H_2O rightleftharpoons HPO_4^(2-) + H_3^(+)O#, #pKa_2 = 7.21#
#HPO_4^(2-) + H_2O rightleftharpoons PO_4^(3-) + H_3^(+)O#, #pKa_3 = 12.7#
Because the pH of blood lies between #pKa_2# and #pKa_3#, the dominant form of the acid will be #HPO_4^(2-)#. Because #pKa_2# and #pKa_3# vary by more than 4 units, the concentration of #PO_4^(3-)# will be negligible.
So, you know that the total phosphate concentration is #"1.05 mmol/L"#. This means that
#[H_2PO_4^(-)] + [HPO_4^(2-)] = 1.05 "mmol/L"# (assuming #[PO_4^(3-)]# is negligible). (1)
Since you're dealing with a buffer, the Henderson-Hasselbalch equation can be used
#[pH](http://socratic.org/chemistry/acids-and-bases/the-ph-concept) = pKa_2 + log(([HPO_4^(2-)])/([H_2PO_4^(-)]))#
#7.40 = 7.21 + log(([HPO_4^(2-)])/([H_2PO_4^(-)])) => ([HPO_4^(2-)])/([H_2PO_4^(-)]) = 1.55#
Plug this value into (1) and you'll get
#([HPO_4^(2-)])/1.55 + [HPO_4^(2-)] = 1.05#, which will result in
#[HPO_4^(2-)] = "0.639 mmol/L"#
The concentration of #H_2PO_4^(-)# will be #[H_2PO_4^(-)] = "0.412 mmol/L"#
If you're interested, you can check the assumption that #[PO_4^(3-)]# is negligible; the equations will produce
#[PO_4^(3-)] = "0.00000320 mmol/L"# #-># for all intended purposes, this is equal to zero.
One more thing...the results match the known proportions of dyhidrogen phosphate and hydrogen phosphate in extracellular fluid (39% for the former, 61% for the latter).
