posted by anonymous .
oxalic acid, h2c2o4, is a weak acid capable of providing two H3O+ ions.
the [h3o]+ in a .38 M solution of h2c2o4 is .12M and can be calculated by the first ionization step only. what is the equilibrium concentration of the oxalate ion c2o4^-2, in the .38 M solution of h2c2o4?
H2C2O4 ==> H^+ + HC2O4^-
HC2O4^- ==> H^+ + C2O4^-2
k2 = (H^+)(C2O4^-2)/(HC2O4^-)
You know (H^+) = 0.12M
For all practical purposes, HC2O4^- is the same; therefore, (C2O4^-2) = k2 = 6.4E-5M.