CH3CH2NH2(aq) + H2O(l) <-> CH3CH2NH3+(aq) + OH-(aq)
Kb = 5.0e-4
Ethylamine, CH3CH2NH, a weak base reacts with water according to the equation above. A 50.0 mL sample of a ethylamine solution is found to have a pH of the solution 11.60.
(a) Write the expression for the equilibrium constant, Kb, for ethylamine.
I got Kb = {[CH3CH2NH3+][OH-]}/[CH3CH2NH2]
(b) Calculate the molar concentration of OH- in the 50.0 mL sample of the ethylamine solution.
I got [OH-] = 3.544e-8 M
(c) Calculate the initial molar concentration of CH3CH2NH2(aq) in the solution before it reacted with water and equilibrium was established.
This one, part c, is the one I'm struggling with.
I think you need to rethink b.
If pH = 11.6 then pH is 2.4 and (OH^-) must be something x E-3.
c. Call ethylamine just BNH2
..........BNH2 + HOH ==> BNH3^+ + OH^-
I..........y..............0........0
C.........-x..............x........x
E.........y-x.............x........x
You know Kb (your expression in part A is correct). You know OH^- (x in the above ICE chart), solve for y.
Well, well, well, looks like you're in a bit of a pickle with part (c). Don't worry, I'm here to help you out!
To solve this, we can use the information given to find the concentration of OH- in the solution and then use the equation for Kb to find the initial molar concentration of CH3CH2NH2.
We already calculated [OH-] to be 3.544e-8 M, so let's plug that into the equation for Kb and solve for [CH3CH2NH2].
Kb = {[CH3CH2NH3+][OH-]}/[CH3CH2NH2]
We know Kb is 5.0e-4 and [OH-] is 3.544e-8 M, so let's plug that in:
5.0e-4 = {[CH3CH2NH3+](3.544e-8)}/[CH3CH2NH2]
Since we're solving for [CH3CH2NH2] initially, let's assume x as the concentration of [CH3CH2NH2] initially.
5.0e-4 = (3.544e-8)(x)
Now, let's solve for x:
x = 5.0e-4 / 3.544e-8
x ≈ 1.413 M
So, the initial molar concentration of CH3CH2NH2 in the solution before it reacted with water and equilibrium was established is approximately 1.413 M.
Hope that helps! And remember, chemistry is all about finding the right balance...just like a clown on a unicycle!
To determine the initial molar concentration of CH3CH2NH2(aq) in the solution, we can use the fact that the solution has a pH of 11.60.
First, we need to find the concentration of H3O+ in the solution using the formula:
pH = -log[H3O+]
11.60 = -log[H3O+]
[H3O+] = 10^(-11.60) M
Since the solution is basic, we can assume that the concentration of OH- ions is equal to the concentration of H3O+ ions. Therefore, [OH-] = 10^(-11.60) M.
Now, we can use the equilibrium expression for Kb to calculate the initial concentration of CH3CH2NH2(aq):
Kb = [CH3CH2NH3+][OH-]/[CH3CH2NH2]
5.0e-4 = ([CH3CH2NH3+])(10^(-11.60))/(initial concentration of CH3CH2NH2)
Let's assume that the initial concentration of CH3CH2NH2(aq) is x M.
5.0e-4 = (10^(-11.60))(10^(-11.60))/x
5.0e-4 = 10^(-23.20)/x
Dividing both sides by 5.0e-4:
1 = (10^(-23.20))/(5.0e-4 * x)
Rearranging the equation:
1 = (10^(-23.20))* (1/(5.0e-4 * x))
Taking the logarithm of both sides:
log(1) = log((10^(-23.20)) * (1/(5.0e-4 * x)))
0 = -23.20 + log(1/(5.0e-4 * x))
Rearranging again:
23.20 = log(1/(5.0e-4 * x))
Converting to exponential form:
10^(23.20) = 1/(5.0e-4 * x)
Simplifying:
x = 1/(10^(23.20) * 5.0e-4)
x = 2.002e-20 M
Therefore, the initial molar concentration of CH3CH2NH2(aq) in the solution before it reacted with water and equilibrium was established is approximately 2.002e-20 M.
To answer part (c), we need to use the pH of the solution and the Kb value to find the initial molar concentration of CH3CH2NH2(aq) before it reacted with water and reached equilibrium.
First, let's convert the pH value to the concentration of H+ ions in the solution. The pH is defined as the negative logarithm of the hydrogen ion concentration. So, we can use the formula:
[H+] = 10^(-pH)
[H+] = 10^(-11.60)
[H+] = 2.51 x 10^(-12) M
Since ethylamine is a weak base, it reacts with water to produce OH- ions. In the equilibrium reaction, every molecule of CH3CH2NH2 reacts with one molecule of H2O to form CH3CH2NH3+ and OH-. This means that at equilibrium, the concentration of OH- ions is the same as the concentration of CH3CH2NH3+.
Using the expression for the equilibrium constant Kb:
Kb = [CH3CH2NH3+][OH-] / [CH3CH2NH2]
we can rearrange the equation to find [CH3CH2NH3+]:
[CH3CH2NH3+] = Kb * [CH3CH2NH2] / [OH-]
Now, substituting the values we know:
Kb = 5.0e-4
[OH-] = 2.51 x 10^(-12) M
[CH3CH2NH3+] = (5.0e-4) * [CH3CH2NH2] / (2.51 x 10^(-12))
We need to find [CH3CH2NH2], the initial molar concentration of ethylamine. We can assume that the initial concentration of [CH3CH2NH3+] is zero since it's the product of the reaction. So, the equation becomes:
[CH3CH2NH3+] = (5.0e-4) * [CH3CH2NH2] / (2.51 x 10^(-12)) = 0
Simplifying the equation:
[CH3CH2NH2] = 0
This means that before reaction, there was no ethylamine present in the solution. It all reacted with water to form the ethylammonium ion (CH3CH2NH3+) and hydroxide ions (OH-).