Calculate the pH of a 0.2M solution of Triethylamine, (C2H5)3N, with a Kb of 4.0x10^-4
..(C2H5)3N + HOH ==> (C2H5)3NH^+ + OH^-
I...0.2M................0...........0
C....-x.................x...........x
E...0.2-x...............x...........x
Substitute the E line into the Kb expression and solve for x = (OH^-), then convert that to H^+ by
(H^+)(OH^-) = Kw = 1E-14
Then pH = -log(H^+)
[OH^-]=8.94x10^-3
pOH=2.05
pH=11.95
[H^+]=1.12x10^-12
To calculate the pH of a solution of Triethylamine, you need to determine the concentration of hydroxide ions (OH-) in the solution, as Triethylamine is a base.
Triethylamine (C2H5)3N is a weak base, so you can use the Kb expression to find the concentration of hydroxide ions.
Kb = [OH-][C2H5)3N] / [(C2H5)3NH+]
First, set up the Kb expression using the given values:
4.0x10^-4 = [OH-][C2H5)3N] / [(C2H5)3NH+]
Since (C2H5)3N is a weak base, we can assume that the concentration of OH- will be equal to the concentration of (C2H5)3NH+. So, we'll have [OH-]^2 in the numerator and [C2H5)3N] in the denominator:
4.0x10^-4 = [OH-]^2 / [C2H5)3N]
We know that the initial concentration of Triethylamine is 0.2M, so we can substitute that value:
4.0x10^-4 = [OH-]^2 / 0.2
Rearranging the equation to solve for [OH-]:
[OH-]^2 = 4.0x10^-4 * 0.2
[OH-]^2 = 8.0x10^-5
Now, take the square root of both sides to find [OH-]:
[OH-] = sqrt(8.0x10^-5)
[OH-] ≈ 0.00894 M
Finally, to find the pH, we need to convert the concentration of hydroxide ions to the concentration of hydrogen ions (H+).
Since water molecules undergo an autoionization reaction, we know that [OH-] x [H+] = 1.0x10^-14 at 25°C.
So, [H+] = 1.0x10^-14 / [OH-]
[H+] = 1.0x10^-14 / 0.00894
[H+] ≈ 1.12x10^-12 M
Taking the negative logarithm of [H+], we can calculate the pH:
pH = -log10([H+])
pH = -log10(1.12x10^-12)
pH ≈ 11.95
Therefore, the pH of a 0.2M solution of Triethylamine is approximately 11.95.
To calculate the pH of a solution of Triethylamine, we need to first find the concentration of hydroxide ions (OH-) in the solution.
Triethylamine, (C2H5)3N, is a weak base. The molecule can accept a proton from water to form hydroxide ions:
(C2H5)3N + H2O ⇌ (C2H5)3NH+ + OH-
The equilibrium constant for this reaction is called the base ionization constant, Kb. In this case, Kb is given as 4.0x10^-4.
The Kb expression can be written as:
Kb = [ (C2H5)3NH+ ] [ OH- ] / [ (C2H5)3N ]
Since we are given the concentration of Triethylamine, which is 0.2M, we can assume that the concentration of hydroxide ions formed is "x". The concentration of (C2H5)3NH+ is also equal to "x" because the mole ratio of (C2H5)3NH+ to OH- is 1:1.
Therefore, we can rewrite the Kb expression as:
(4.0x10^-4) = ( x ) ( x ) / (0.2)
Now, solve for "x":
4.0x10^-4 = x^2 / 0.2
Multiply both sides by 0.2:
0.2 * (4.0x10^-4) = x^2
8.0x10^-5 = x^2
Take the square root of both sides:
x = √(8.0x10^-5)
x ≈ 0.00894
Now that we have the concentration of hydroxide ions (OH-), we can calculate the concentration of hydrogen ions (H+). Since the solution is neutral, the concentration of H+ is equal to the concentration of OH-.
Therefore, the pH of the solution can be calculated using the equation:
pH = -log[H+]
pH = -log(0.00894)
pH ≈ 2.05
So, the pH of a 0.2M solution of Triethylamine with a Kb of 4.0x10^-4 is approximately 2.05.