How to finish your lab report for the pKa of HCu(C2 O4 )2 (H2 O)(OH)2- (aq).
Recall that we agreed to refer to the acid HCu(C2O4)2(H2O)(OH)2- as HA. The figure below
shows a titration curve something like yours. The reaction that takes place during the titration
is:
HA + OH-  H2O + ATo summarize: 1. Before the titration begins, you have almost 100% HA in the flask. As OHwas added, the HA decreased and the A- increased. Below the curve below, I have indicated
how HA decreases from 100% to 0% by the time we get to the equivalence point. Meanwhile,
the amount of A- increases from 0% to 100%. In other words, the stuff is either HA or A-.
Note in green I have indicated what is present ¼ of the way to the equivalence point.
See the pre-lab notes you recorded during the pre-lab lecture: We use the data for the first
derivative graph to precisely determine the equivalence point. The graph shows you the
approximate equivalence point and then looking at the first derivative data to find the precise
maximum – and the volume that goes with it. Then you divide this volume by 2 to get the “halfway to the equivalence point” volume. Find this volume in the data table and the
corresponding pH (from the data table) is the pKa! Simple as that!
Hints for the post-lab questions:
1. Define molarity, use the mass of the blue crystals measured out. Why must you use the
molar mass for K2[Cu(C2O4)2(H2O)2]?
2. Think about the balanced chemical equation that takes place between a weak acid and a
strong base.
5. MMHA = mHA/nHA. The mass part is easy. For nHA, remember that at the equivalence
point (and only at the equivalence point), nHA = nOH-. The latter is obtained from nOH- =
MOH-VOH-