Chemistry

1. Calculate the concentration of H3O+, HC2O4-1, and oxalate ion (C2O42-) in a 0.175 M solution of oxalic acid (C2H2O4).
[For oxalic acid, Ka1 = 6.5 ´ 10-2, Ka2 = 6.1 ´ 10-5.]

asked by Anonymous
  1. .......H2C2O4 + H2O --> H3O^+ + HC2O4^-
    I......0.175.............0.......0
    C........-x..............x.......x
    E.....0.175-x............x.......x

    Substitute the E line into Ka1 expression and solve for x = (H3O^+) = (HC2O4^-).

    That gives you H3O^+ and HCO4^- and note they are equal. It also gives (H2C2O4). That leaves C2O4^2-


    .......HC2O4^- + H2O ==> H3O^+ + C2O4^2-
    Ka2 = (H3O^+)(C2O4^2-)/(HC2O4^-)
    Since (H3O^+) = (HC2O4^-), then (C2O4^2-) = ka2.

    posted by DrBob222

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