Asked by Paul
Kb for aniline is 4.3 x 10^-10. What does this tell you about aniline? Please help!
Answers
Answered by
DrBob222
It tells you it is a weak base.
If we call aniline phNH2 (ph is for C6H5), then in water solution,
phNH2 + HOH ==> phNH3^+ + OH^-
Kb = (phNH3^+)(OH^-)/(phNH2)
Write the Kb and this equation on paper. Kb = 4.3 x 10^-10.
Note that for that fraction to be less than 1 (and 4.3 x 10^-10 is WAY less than 1), the products (in the numerator) MUST be MUCH smaller than the reactants (in the denominator) so it means that there is very little phNH3^+ and very little OH^- and that most of the material is in the form of aniline. That means that the reaction occurred to a very small extent.
If we call aniline phNH2 (ph is for C6H5), then in water solution,
phNH2 + HOH ==> phNH3^+ + OH^-
Kb = (phNH3^+)(OH^-)/(phNH2)
Write the Kb and this equation on paper. Kb = 4.3 x 10^-10.
Note that for that fraction to be less than 1 (and 4.3 x 10^-10 is WAY less than 1), the products (in the numerator) MUST be MUCH smaller than the reactants (in the denominator) so it means that there is very little phNH3^+ and very little OH^- and that most of the material is in the form of aniline. That means that the reaction occurred to a very small extent.
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