# How do you find conjugate base concentration?

Mar 31, 2014

The easiest way is to use the Henderson-Hasselbalch equation.

EXAMPLE:

The $\text{p} {K}_{a}$ for the dissociation of ${\text{H"_3"PO}}_{4}$ is 2.15. What is the concentration of its conjugate base ${\text{H"_2"PO}}_{4}^{-}$ at pH 3.21 in 2.37 mol/L phosphoric acid?

Solution:

The equation is

${\text{H"_3"PO"_4 + "H"_2"O" ⇌ "H"_3"O"^+ + "H"_2"PO}}_{4}^{-}$

For simplicity, let’s rewrite this as

${\text{HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A}}^{-}$

The Henderson-Hasselbalch equation is:

"pH" = "p"K_"a" + log((["A"^-])/("[HA"]))

Substituting,

$3.21 = 2.15 + \log \left(\left(\left[\text{A"^-])/(["HA}\right]\right)\right)$

Solving,

log((["A"^-])/(["HA"])) = 3.21 – 2.15 = 1.06

$\left(\left[\text{A"^-])/(["HA}\right]\right) = {10}^{1.06} = 11.5$, or

$\left[\text{A"^-] = 11.5["HA}\right] .$

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".