How do you find the Ka1 and Ka2 of oxalic acid when it is in a solution that is 1.05 M H2C2O4 and has a pH of 0.67. [C2O4^2-] = 5.3x10^-5 M. I have tried using an ICE table for both reactions, as oxalic acid is a diprotic acid, but I am having trouble knowing where to put concentrations on these tables, and where to put unknowns. I have found [H3O+] to be 0.214 M, but am not sure where to go from there. The reactions are H2C2O4 + H2O -- H3O^+ + HC2O4^- (Ka1 = ?) and HC2O4^- + H2O -- H3O^+ + C2O4^2- (Ka2 = ?)

Question

How do you find the Ka1 and Ka2 of oxalic acid when it is in a solution that is 1.05 M H2C2O4 and has a pH of 0.67. [C2O4^2-] = 5.3x10^-5 M. I have tried using an ICE table for both reactions, as oxalic acid is a diprotic acid, but I am having trouble knowing where to put concentrations on these tables, and where to put unknowns. I have found [H3O+] to be 0.214 M, but am not sure where to go from there. The reactions are H2C2O4 + H2O --> H3O^+ + HC2O4^- (Ka1 = ?) and HC2O4^- + H2O --> H3O^+ + C2O4^2- (Ka2 = ?)

bobpursley · Answer

You know the concentration of the chromate ion (above), then let the concentration of the romate ion be x, then x+concentrationroamte=1.05m

It seems to me then you can figure out the H+ ion concentration in each reaction, and thence figure Ka for each. I didn't work it out, but it seems straightforward to my simplistic mind.