chemistry

oxalic acid, h2c2o4, is a weak acid capable of providing two H3O+ ions.
ka=.059
k2=6.4E-5
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?

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  1. 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.

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    posted by DrBob222

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