Oxalic acid is a diprotic acid that occurs naturally in some plants. Calculate t
ID: 726178 • Letter: O
Question
Oxalic acid is a diprotic acid that occurs naturally in some plants. Calculate the pH and the concentrations of all species present in a 0.36-M solution. The acid dissociation constants are K1 = 5.9 10-2 and K2 = 6.4 10-5.a) [C2O4H2] .24 M
b) [C2O4H ? ] .12 M
c) [C2O42-] M
d) [H+] M
e) pH
I got Parts a and b, but can't get the rest even though I'm pretty sure you just have to do the same step again but with different values :(
Explanation / Answer
the equilibrium equation for the dissociation HC3H5O2 ? C3H5O2- + H+ is Ka = [H+] [C3H5O2-] / [HC3H5O2] Initially, [H+] = 0, [C3H5O2-] = 0, [HC3H5O2] = 0.207 At equilibrium, [H+] = x, [C3H5O2-] = x, [HC3H5O2] = 0.207-x So Ka = x² / (0.207-x). Assuming 0.207>>x, the equation rearranges to: x = v((0.207)(1.3x10^-5)) = 0.00164 The answers: [H+] = x = 0.00164 M pH = -log[H+] = -log(0.00164) = 2.79 [C3H5O2-] = x = 0.00164 M [-OH] = (10^-14) / [H+] = (10^-14) / (0.00164) = 6.08x10^-12 M [HC3H5O2] = 0.207-x = 0.207 M - 0.00164 M ˜ 0.205 M Percent dissociation = [C3H5O2-] / [HC3H5O2] = 0.00164 M / 0.205 M = 0.0078, or about 0.8%.
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