Question

Ethylamine, CH3CH2NH2, has a strong, pungent odor similar to that of ammonia. Like ammonia, it is a Brønsted base. A 0.10 M solution has a pH of 11.87. Calculate the Kb and pKb for ethylamine. What is the percentage ionization of ethylamine in this solution?

Answers

Let's write ethylamine as BNH2. If the pH is 11.87 the pOH is 14-11.87 = pOH of 2.13 and 2.13 = -log(OH^-) so (OH^- is approx 7E-3M but you need to that more accurately as well as all of the other calculations that follow. Then
.......BNH2 + HOH ==> BNH3^+ + OH^-
I......0.1.............0........0
C.......-x.............x........x
E......0.1-x...........x........x
and you know x = approx 0.007 so 0.1-x = approx 0.093. Plug those recalculated values into the Kb expression and solve for Kb. Then pKb = -logKb
% ion = [(OH^-)/0.1]*100 = ?
Kb = (BNH3^+)(OH^-)/(BNH2)

Ethylamine (CH3CH2NH2), has a strong pungent odour similar to that
of ammonia. Like ammonia it is a Bronsted base. A 0.10 M solution
of ethylamine has a pH of 11.86. Calculate the Kb and pKb for
ethylamine.
The first step is to write the equilibrium reaction for the base dissociation of ethylamine:

CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH-

From the pH of the solution, we can determine the [OH-] concentration:

pH + pOH = 14
pOH = 14 - pH = 14 -11.86 = 2.14
[OH-] = 10^(-pOH) = 10^(-2.14) = 6.97 x 10^(-3) M

Next, we can set up an ICE table for the reaction:

CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH-
I 0.10 M 0 M 0 M
C -x x x
E 0.10 - x x x

Using the equilibrium concentrations in the ICE table, we can write the expression for the base dissociation constant (Kb):

Kb = [CH3CH2NH3+][OH-]/[CH3CH2NH2]

Substituting the values from the ICE table and solving for Kb:

Kb = (x)(x)/(0.10 - x)
Kb = 5.6 x 10^-4

Finally, we can calculate the pKb:

pKb = -log(Kb) = -log(5.6 x 10^-4) = 3.25

Therefore, the Kb for ethylamine is 5.6 x 10^-4 and the pKb is 3.25.

Related Questions