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Acidity of isobutane

  1. Nov 17, 2015 #1
    1. The problem statement, all variables and given/known data

    Don't want to give too many details, but a homework question gives a pKa value for isobutane (CH3)3CH as 71. I'm obviously missing something because when I try to calculate Kb out of curiosity I get an astoundingly large value (especially considering I don't even see how isobutane could be a base!).

    2. Relevant equations

    Ka = 10^(-pKa)

    Kw = Ka * Kb

    3. The attempt at a solution

    The question has the isobutane in a relatively dilute (1e-3 M) solution. I calculated (using the equations above) Ka to be 1e-71, an extremely weak acid which makes sense to me, but then if I use that value in the second equation, I get Kb = Kw/Ka = 1e57. That can't be right! How would isobutane even function as a base, not to mention that's a ridiculously enormous value of Kb. What am I missing? Thanks!
  2. jcsd
  3. Nov 17, 2015 #2


    User Avatar

    Staff: Mentor

    Kw=Ka*Kb works for water solutions, and even then not for very weak and very strong acids.

    Very strong acid will simply fully dissociate protonating water to H3O+ and H3O+ will be the only acid present - so it is the strongest possible acid in water solutions. This is called "leveling effect". Similarly, no base can be stronger than OH-. For a very weak acid (like isobutane) presence of a much stronger acid (water itself) means its pKa doesn't matter (in water solutions).

    Such very high values of pKa/pKb can be used to predict what happens in organic chemistry, when substances react in other solvents in water. While they help give some qualitative predictions, this is quite handwavy.
  4. Nov 17, 2015 #3
    Perfect. Thanks so much!
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