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?

Answer 1

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)

Answer 2

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.

Answer 3

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.