# How to draw distribution graph if pKa of acid is 4.4 and pKa of base is 6.7?

## it should look like this

Jul 4, 2017

Well, these distribution graphs should correlate with the titration curve.

If we know the first $\text{pKa}$ is $4.4$ and the second $\text{pKa}$ is $6.7$, then we have an idea of where the half-equivalence points are (i.e. where the concentrations of acid and conjugate base are equal), because the $\text{pH}$ $=$ $\text{pKa}$ at those points:

"pH"_("1st half equiv. pt.") = "pKa"_1 + cancel(log\frac(["HA"^(-)])(["H"_2"A"]))^("Equal conc.'s, "log(1) = 0)

"pH"_("2nd half equiv. pt.") = "pKa"_2 + cancel(log\frac(["A"^(2-)])(["HA"^(-)]))^("Equal conc.'s, "log(1) = 0)

We represent each stage of a diprotic acid as:

${\text{H"_2"A"(aq) rightleftharpoons overbrace("HA"^(-)(aq))^"singly deprotonated" + "H}}^{+} \left(a q\right)$

rightleftharpoons overbrace("A"^(2-)(aq))^"doubly deprotonated" + "H"^(+)(aq)

The two midpoints shown are the first and second half-equivalence points, respectively.

• At midpoint 1, we have that $\left[{\text{H"_2"A"] = ["HA}}^{-}\right]$, and that $\text{pH} \approx 4.4$.

• At midpoint 2, we have that $\left[{\text{HA"^(-)] = ["A}}^{2 -}\right]$, and that $\text{pH} \approx 6.7$.

A distribution graph shows the change in concentration of each species in solution as the $\text{pH}$ increases. It correlates well with a base-into-diprotic-acid titration curve.

See below for an overlay of both:

Each species in solution is tracked in the bottom graph.

• The cross-over points on the distribution graph are the half-equivalence points on the titration curve.

• The maximum concentration for each species after the starting $\text{pH}$ correlate with the first few equivalence points, and the last species to show up dominates at high $\text{pH}$.