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Chemistry

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Calculate [CO32- ] in a 0.019 M solution of CO2 in water (H2CO3). If all the CO32- in this solution comes from the reaction shown below, what percentage of the H+ ions in the solution is a result of the dissociation of HCO3‾?
HCO3‾(aq) H+(aq) + CO32-(aq)

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    ........H2CO3 --> H^+ + HCO3^-
    initial.0.019....0........0
    change...-x.......x.......x
    equil.0.019-x.....x........x

    k1 = (H^+)(HXO3^-)/(H2CO3).
    Substitute into k1 expression and solve for H^+. I obtained approximately 1E-4 but you should do it more accurately.

    Then HCO3^- ==> H^+ + CO32^2-
    init.1E-4.......0.......0
    change.-x.......x.......x
    equil...1E04-x...x.......x

    I have show the two above as if the H^+ from equation 1 is not the same as in equation 2. Actually, they are the same for both equation; however, I have done it this way to show the different.
    k2 = (H^+)(CO3^2-)/(HCO3^2-)
    But from equation 1 we see that (H^+) = (HCO3^-), which makes from equation 2 that (C03^2-) = k2 = about 5E-11

    So what percentage of the H^+ comes from k2? Very small.
    [(H^+)from k2/(H^+)from k1]*100 =% from the second ionization.

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