How to draw distribution graph if pKa of acid is 4.4 and pKa of base is 6.7?
it should look like this
it should look like this
1 Answer
Well, these distribution graphs should correlate with the titration curve.
If we know the first
#"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:
#"H"_2"A"(aq) rightleftharpoons overbrace("HA"^()(aq))^"singly deprotonated" + "H"^(+)(aq)#
#rightleftharpoons overbrace("A"^(2)(aq))^"doubly deprotonated" + "H"^(+)(aq)#
The two midpoints shown are the first and second halfequivalence points, respectively.

At midpoint 1, we have that
#["H"_2"A"] = ["HA"^()]# , and that#"pH" ~~ 4.4# . 
At midpoint 2, we have that
#["HA"^()] = ["A"^(2)]# , and that#"pH" ~~ 6.7# .
A distribution graph shows the change in concentration of each species in solution as the
See below for an overlay of both:
Each species in solution is tracked in the bottom graph.

The crossover points on the distribution graph are the halfequivalence points on the titration curve.

The maximum concentration for each species after the starting
#"pH"# correlate with the first few equivalence points, and the last species to show up dominates at high#"pH"# .