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

it should look like this

enter image source here

1 Answer
Jul 4, 2017

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

If we know the first #"pKa"# is #4.4# and the second #"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 #"pH"# #=# #"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)#

https://files.mtstatic.com/

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 half-equivalence 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 #"pH"# increases. It correlates well with a base-into-diprotic-acid titration curve.

See below for an overlay of both:

Titration Curve (Truong-Son N.) + Distribution Graph (Ernest Z.)

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 #"pH"# correlate with the first few equivalence points, and the last species to show up dominates at high #"pH"#.