acids/bases Q (general chemistry)

Discussion in 'MCAT Study Question Q&A' started by crazymedgirl, May 17, 2008.

  1. crazymedgirl

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    Hi, was wondering if anyone out there could help me with this acids/bases problem I'm stumped on:

    The equilibrium constant for dissolving CaCO3 in water is 3.4E-9. What is the equilibrium constant for dissolving CaCO3 in the presence of acid? (Ka of H2CO3 = 4.2E-7; Ka of HCO3- = 4.8E-11)

    a. 1.6E-19
    b. 2.9E8
    c. .008
    d. 70.8

    The correct answer is d.

    I am confused about how to even approach this problem, or make use of the information being given. Also, the Ka for CO3^2- is 2.1E-4, not 3.4E-9, so I'm not even sure where that number comes from...

    Any help would be greatly appreciated.
     
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  3. BerkReviewTeach

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    First and foremost, where did you get this question? Is this from a chemistry course, because it doesn't have all of the information present? It has a few issues besides being beyond the scope of the MCAT.

    1) CO32- is a base with NO proton to lose, so it CANNOT have a Ka. Its Kb is 2.1exp-4, which is found because Ka x Kb = 10exp-14.

    2) To truly solve it, you need to know the Ka for the acid they are using. for instance, if it's strong acid, the overall reaction will be favorable. If it's a weak acid, then the overall reaction will be unfavorable (less than 1). You really need to know the acid being used.

    3) Once you add acid, it becomes a complex equilibrium, and the final products are not presented in the question. Did they mention that CO2 was evolved?

    So, with that in mind, let's consider the four reactions that will sum to the overall reaction.

    CaCO3 <=> Ca2+ + CO32-

    and

    2 HX <=> 2 H+ + 2 X-

    CO32- + H+ <=> HCO3-

    HCO3- + H+ <=> H2CO3

    H2CO3 <=> H2O + CO2

    The overall reaction is:

    CaCO3 + 2 H+ <=> Ca2+ + 2 X- + H2O + CO2

    The Keq for dissolving CaCO3 in acid is the product of the Keqs for the reactions being summarized.

    Now assuming they want the final product to be Ca2+ and H2CO3, then the Keq would be found by multiplying the Keqs for the first four reactions listed above.

    (3.4 x 10exp-9) X (2.1 x 10exp-4) X (2.4 x 10exp-8) X (Ka)exp2 = 16.2 x 10exp-21 X (Ka)exp2 = 1.62 x 10exp-20 X (Ka)exp2.

    It can be solved once we know the Ka for the acid they used.
     
  4. crazymedgirl

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    thanks a lot for taking the time to respond. yes, it's actually from an Honors General Chemistry course. I'm not sure why it's at such an unusually high level....

    thanks again.
     

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