Explore BrainMass

Explore BrainMass

    Diprotic acid with ka1 and ka2

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    A diprotic acid, H2A, has the following ionization constants: Ka1 = 1.1E-3 and Ka2 + 2.5E10-6. In order to make up a buffer solution of pH 5.80, which combination would you choose, NaHA/H2A or Na2A/NaHa? What is pKa of the acid component?
    I don't understand this problem at all! I know that a diprotic substance has something in common in each step: H2A--->H+ + HA- and HA- ----->H+ + A2-.
    What exactly is pKa? I believe its just the -log(Ka). Does "p" somehow stand for -log such as pOH=-log[OH-].
    If possible explain in simple steps. Thank-you.

    © BrainMass Inc. brainmass.com March 4, 2021, 6:19 pm ad1c9bdddf
    https://brainmass.com/chemistry/acids-and-bases/diprotic-acid-ka1-ka2-39978

    Solution Preview

    You're quite right about pKa being -log(Ka) but this has an important implication. Follow this argument...

    Ka = [H+][Acid-]/[HAcid]

    Note that in your case, the HAcid/Acid- pairs are H2A and HA- in the first case, and HA- and A2- in the second case. Anyway, the equation can be rearranged to:

    [H+] = Ka[HA]/[A-]

    This is very useful for buffer solutions because we can calculate [H+] (and hence pH) if we know Ka and ...

    Solution Summary

    The PKa fir log(Ka) are found. The question is answered step-by-step equationally with explanations throughout.

    $2.49

    ADVERTISEMENT