- #1
sciencegem
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Homework Statement
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!).
Homework Equations
Ka = 10^(-pKa)
Kw = Ka * Kb
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!