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".
