Question

Problem: Need to prepare a 250 mL buffer of pH 7.00 using equimolar K2HPO4 and KH2PO4 solutions. Ka for phosphoric acid:
Ka1=7.2 x 10^-3
Ka2=6.3 x 10^-8
Ka3=4.2 x 10^-13

I though the hard part of this was picking the right equation/Ka value, but I am not getting the right answer, so not sure what I else I may be doing wrong.

So equation used:
H2PO4(-) + H2O <-->HPO4(2-) + H3O(+) (Ka2 = 6.3 x 10^-8)
[weak acid] [conjugate base]

K+ are spectator ions

[going to drop the charges for the rest--it is too confusing in this font]
Ka2 = [HPO4][H3O] / [H2PO4]
Desired pH = 7.00, so [H3O] =10^-7, so
Ka/[H3O] = (6.3 s 10^-8)/10^-7 = 0.63 = [HPO4]/[H2PO4]

Both solutions have same molarity, so volume ratio should match mole ratio so HPO4 vol + H2PO4 vol = 250 mL = 0.63 H2PO4 +1 H2PO4, so
H2PO4 vol =250 mL/1.63 = 153.4 mL KH2PO4
HPO4 vol = 250 mL -153.4 mL =96.6 mL K2HPO4

Do I even have the right approach? Am I missing something I should be factoring in?

Thanks much.

Answer 1

I don't see anything wrong with this. I would have used the HH equation but what you used is perfectly ok. In fact, if you substitute those values into the HH equation you get a pH of 6.999 which should be ok.
pH = 7.20 + log(96.6/153.4) = 6.999 which rounds to 7.0 and that's was you wanted.
If I had to hazard a guess, I would think the trouble is not with how yu worked it but the fact that you are reporting too many significant figures. What happens if you round that to 97/153 for base/acid as an answer. The pH by the HH equation is 7.00 on the button and you are using only 3 s.f.