Acidity of Isobutane: pKa Value & Kb Calculation

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SUMMARY

The pKa value of isobutane ((CH3)3CH) is established at 71, indicating it is an extremely weak acid. The calculation of Kb using the formula Kb = Kw/Ka results in an impractically high value of 1e57, which raises questions about isobutane's ability to act as a base. The discussion highlights the "leveling effect" in water, where very weak acids like isobutane are overshadowed by the stronger acid, water, making their pKa values less relevant in aqueous solutions. This emphasizes the limitations of using pKa and pKb values for predicting behavior in organic chemistry.

PREREQUISITES
  • Understanding of acid-base chemistry, specifically pKa and Kb calculations.
  • Familiarity with the leveling effect in aqueous solutions.
  • Knowledge of the relationship between Ka, Kb, and Kw.
  • Basic principles of organic chemistry reactions in various solvents.
NEXT STEPS
  • Research the concept of the leveling effect in more detail.
  • Explore the implications of pKa and pKb values in non-aqueous solvents.
  • Study the behavior of weak acids and bases in organic chemistry contexts.
  • Learn about the calculation and significance of Ka and Kb for various compounds.
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Chemistry students, organic chemists, and educators seeking to deepen their understanding of acid-base behavior and the limitations of pKa and Kb values in aqueous solutions.

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!
 
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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.
 
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Borek said:
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.
Perfect. Thanks so much!
 

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