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Chem

Discussion in 'MCAT Discussions' started by killinsound, May 11, 2007.

  1. killinsound

    Physician

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    This chemistry question is driving me nuts.

    To make a solution, 7.38 mmol of a diprotic acid (quinolinic acid), is dissolved in 162.5 mL of water. The pH of the resulting solution is measured to be 1.99. From the equation, what is the pKa of the acid?


    i am using henderson hasselbach

    pka = ph - log([A-]/[HA])

    [H2A] = 7.38/162.5 = .0454
    ph = 1.99
    [H+]/2 = [A-]
    (10^-1.99)/2 = .005116

    plug and chug... i get 2.93

    the answer is 2.53.

    what am i doing wrongggggggg?
     
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  3. CATallergy

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    I'm not good at following other people's work, so I'll just solve from beginning:

    1.99 pH means [H+] = 10^-1.99 = .0102

    [H2A] = 7.38 / 162.5 = .0454

    dissociation reaction:
    [H2A] -> [H+] + [HA-]

    Ka = [H+][HA-]/[H2A]

    let x = [H+];

    Ka = x^2 / (.0454 - x)

    since x = .0102,

    (.0102)^2 / (.0454 - .0102) = .002955

    -log .002955 = 2.529
     
  4. scottj72

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    I think because they mentioned the acid was diprotic it really gets you to thinking. Because actually a lot more going on, so it gets distracting.
    H2A --> H+ + HA-
    HA --> H+ + A-
    But since they give the pH you know the total concentration of [H+] dissociated. 1.99 pH means [H+] = 10^-1.99 = .0102
    Even if the acid is diprotic the net ratio of the concentrations is 1:1.
    We do not know all the exact concentrations be we do have enough to determine the net concentrations.
    because {[H+] + [H+]} = {[HA-] + [A-]} = .0102
    [H2A] = total moles - dissociated moles

    so pKa = pH - log {[HA-] + [A-]} / [H2A]

    pKa = 1.99 - log (.0102 / [.0454 - .0102] )
    pka = 1.99 - log (0.2898)
    pKa= 2.528
     
  5. BrokenGlass

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    quinolinic acid is a weak acid (because it's not on the list of known strong acids)

    pH = (pKa + pH (if strong))/2 (This is a really good shortcut equation to know; it calculates pH for an aqueous solution of a weak acid; there are restrictions on its use; I don't remember them now; lets assume they are met; I'll post a correction if the are not)

    pH (if strong) = -log[H+] = -log[quinolinic acid] = -log[7.38/162.5] = -log[0.0454] = 1.3428

    1.99 = (pKa + 1.3428)/2
    3.98 = pKa + 1.3428
    pKa = 2.63

    Since pH = (pKa + pH (if strong))/2 is just an approximation, the answer obtained using this formula would be a little off.
     

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