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"#.