chemistry

For 6.0*10^-2 M H2CO3, a weak diprotic acid, calculate the following values. Use ionization constants of H2CO3: Ka1=4.4*10^-7, Ka2=4.7*10^-11, as necessary.
PART A: [H3O^+] M
PART B: [HCO3^-] M
PART C: [CO3^2-] M

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asked by aliceallyx3
  1. A.
    ........H2CO3 + H2O ==> H3O^+ + HCO3^-
    I.......0.06..............0......0
    C........-x...............x.......x
    E......0.06-x..............x......x

    Substitute the values from the ICE chart into ka1 and solve for x = (H3O^+) = (HCO3^-)

    I've answered part B in part A answer.
    C.
    ......HCO3^- + H2O ==> H3O^+ + CO3^2-
    Note ka2 = (H3O^+)(CO3^2-)/(HCO3^-)
    Write that out on a sheet of paper so you can see it. From part A you know (H3O^+) = (HCO3^-). Therefore, )H3O^+) in the numerator cancels (HCO3^-) in the denominator and (CO3^2-) = ka2.

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    posted by DrBob222

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