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?
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.
To calculate the Kb (base dissociation constant) for ethylamine, we can use the pH of the solution. The Kb of a weak base can be calculated using the formula:
Kb = (Kw / Ka)
Where Kw is the ion product of water and Ka is the acid dissociation constant. Since ethylamine is a base, we need to calculate the Kb using the pOH of the solution and then convert it to pKb.
1. Calculate pOH:
pOH = 14 - pH
= 14 - 11.87
= 2.13
2. Calculate Kb using pOH:
pKb = 14 - pOH
= 14 - 2.13
= 11.87
So, the pKb for ethylamine is 11.87.
3. Calculate Kb from pKb:
Kb = 10^(-pKb)
= 10^(-11.87)
Now, let's move on to calculating the percentage ionization of ethylamine in the solution.
The percentage ionization of a weak base can be given by the formula:
% Ionization = (Concentration of ionized base / Initial concentration of base) * 100
In this case, we have a 0.10 M solution of ethylamine and want to find the percentage ionization.
4. Let x be the concentration of ionized base:
% Ionization = (x / 0.10) * 100
Since ethylamine is a weak base, it partially ionizes to form ethylammonium ion (CH3CH2NH3+) and hydroxide ion (OH-). Since both the base and its conjugate acid are present in equal amounts at equilibrium, we can assume that [OH-] = [CH3CH2NH3+].
5. To calculate [OH-]:
pOH = -log[OH-]
2.13 = -log [OH-]
Solving for [OH-]:
[OH-] = 10^(-2.13)
Since [OH-] = [CH3CH2NH3+], we can substitute this value into the equation for % ionization:
% Ionization = (10^(-2.13) / 0.10) * 100
Calculating this will give you the percentage ionization of ethylamine in the solution.
Note: The calculated values provided above are just an example. To get precise values, you may need to use more accurate experimental data or specific concentration values.
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)