Question

Repost:: sorry couldn't understand it
What is the concentration of C2O4^-2 in a 0.370M oxalic acid solution? Ka1=5.6x10^-2 and Ka2=5.1x10^-5

Answer 1

This is what Dr. Bob222 gave you as an answer in your previous post:


Here is what I would do.
.......H2C2O4 ==> H^+ + HC2O4^-
I.......0.370......0.......0
C........-x.........x......x
E......0.370-x......x......x

BUT there is no need to solve the equation. We simply note that (H^+) = (HC2O4^-) and move on to k2.

k2
.......HC2O4^- ==> H^+ + C2O4^2-
I.....x(from above)..0.....0

k2 = (H^+)(C2O4^2-)/(HC2O4^-)
BUT from k1 we saw (H^+) = (HC2O4^-); therefore, H^+ in the numerator cancels with HC2O4^- in the denominator so that k2 = (C2O4^2-)

You can do this for several reasons. Note that we assumed the total H^+ = just H^+ from k1. We can do that because the H^+ supplied by k2 is 1000 times less than that supplied by k1. And we can assume HC2O4^- we used for k2 is the same as (H^+) because only 1/1000 of those HC2O4^- ions will ionize further.