The easiest way is to use the Henderson-Hasselbalch equation.
EXAMPLE:
The #"p"K_a# for the dissociation of #"H"_3"PO"_4# is 2.15. What is the concentration of its conjugate base #"H"_2"PO"_4^-# at pH 3.21 in 2.37 mol/L phosphoric acid?
Solution:
The equation is
#"H"_3"PO"_4 + "H"_2"O" ⇌ "H"_3"O"^+ + "H"_2"PO"_4^-#
For simplicity, let’s rewrite this as
#"HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^-#
The Henderson-Hasselbalch equation is:
#"pH" = "p"K_"a" + log((["A"^-])/("[HA"]))#
Substituting,
#3.21 = 2.15 + log((["A"^-])/(["HA"]))#
Solving,
#log((["A"^-])/(["HA"])) = 3.21 – 2.15 = 1.06#
#(["A"^-])/(["HA"]) = 10^1.06 = 11.5#, or
#["A"^-] = 11.5["HA"].#
Since the original concentration was 2.37 mol/L,some re-formatting plus a:
#["A"^-] + ["HA"] = "2.37 mol/L"#.
Substituting,
#11.5["HA"] + ["HA"] = (11.5 + 1)["HA"] = 12.5["HA"] = "2.37 mol/L"#.
Solving,
#["HA"] = (2.37" mol/L")/12.5 = "0.190 mol/L"#.
Thus,
#["A"^-] = ["H"_2"PO"_4^-] = ("2.37 - 0.190) mol/L" = "2.18 mol/L"#.
