Calculate the [H3O+] of the following polyprotic acid solution: 0.135 M H2CO3 .?

If someone could please help me it would be much appreciated. I could really use step by step instructions because I can’t seem to figure out how to get the proper Ka.

1 Answer

  • If the Ka or pKa values are not provided in the question ( they should be) then you have to source these from some reference list – I often find some slight differences depending on the source. So final answers may differ slightly between different contrbutors. But the calculation theory and methods should be the same.

    From my reference list I find:

    Ka1 H2CO3 = 4.3*10^-7, pKa = 6.37

    Ka2 H2CO3 = 5.6*10^-11, pKa = 10.25

    Because H2CO3 is a diprotic weak acid, we will make some assumptions that will not affect the end result, but will allow much easier calculation:

    H2CO3 dissociates in two steps:

    H2CO3 ↔ H+ + HCO3-

    HCO3- ↔ H+ + CO3 2-

    We will assume:

    1) [H+] = [HCO3-]= [CO3 2-]

    [H2CO3] = [HCO3-]

    For first dissociation:

    Ka = [H+]*[HCO3-] / [H2CO3]

    Ka = [H+]²/[H2CO3]

    4.3*10^-7 = [H+]²/ 0.135

    [H+]² = 5.805*10^-8

    [H+] = 2.41*10^-4

    For second dissociation:

    Ka = [H+]*[CO3 2-]/[HCO3-]

    5.6*10^-11 = [H+]²/[HCO3-]

    5.6*10^-11 * 0.135 = [H+]²

    [H+]² = 7.56*10^-12

    [H+] = 2.75*10^-6

    Total [H+] ( or [H3O+] = (2.41*10^-4) + 2.75*10^-6) = 2.437*10^-4M

Leave a Comment