Given a weak base such as $\ce{C2H5NH2}$, I usually see the expression for $K_\text{b}$ as

$K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{OH-}]}{[\ce{C2H5NH2}]}$.

Are the expressions

$K_\text{b} = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$, $K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$

equivalent? Do they represent the same process (if so, why do the latter two not work with calculations?) or fundamentally different ways of dissociation? Which one represents what's happening molecularly most accurately? Given the Brønsted-Lowry definition of a base as a proton acceptor I'd think that the second one is most accurate, but I rarely see it used.


2 Answers 2


I try to say it clearer and simpler.

The second expression respects to the equilibrium

$\ce {C2H5NH2 + H+ <=> C2H5NH3+}$.

And the third

$\ce {C2H5NH2 + H3O+ <=> C2H5NH3+ + H2O}$.

However, what you want to know is the equilibrium constant of

$\ce {C2H5NH2 + H2O <=> C2H5NH3+ + OH-}$.

The former two both describe how $\ce {C2H5NH3+}$ behaving as a weak acid in aqueous solution, and the third one describes how $\ce {C2H5NH2}$ behaving as a weak base in aqueous solution (hydrolyzation). They are different because in the former two equilibrium $\ce {C2H5NH2}$ are not receiving proton directly from $\ce {H2O}$.

Or you may look it by this view: the last one can be seen as the result of two equilibrium:

  1. $\ce {C2H5NH2 + H+ <=> C2H5NH3+}$
  2. $\ce {H2O <=> H+ + OH-}$

Or you may write as

  1. $\ce {C2H5NH2 + H3O+ <=> C2H5NH3+ + H2O}$
  2. $\ce {2H2O <=> H3O+ + OH-}$

So you can see there is obvious different. The equilibrium constant of $\ce {2H2O <=> H3O+ + OH-}$ is not 1, so the base dissociation constant $K_b$ will not be the same as what you said.

And there is an important convention that in an aqueous solution system, the equilibrium constant should omit [$\ce {H2O}$] because it is an insignificant constant in the most cases.


$$K_\mathrm b = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$$ This expression is infact $1/K_\mathrm a$ which is related with $K_\mathrm a K_\mathrm b=K_\mathrm w$

$$K_\mathrm b = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$$ here $K_\mathrm b$ is correct but nowadays any constant concentration (here water) is not written in the expression. As it is assumed that concentration of water is not changing when reaction proceed.


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