Calculate [OH − ], [H+], and the pH of a 0.12 M solution of each of the following amines. Calculate it for Aniline
Aniline is a weak base, so we can calculate the concentration of hydroxide ions ([OH-]) using the equation for the base ionization constant (Kb) of aniline:
Kb = [OH-][C6H5NH+]/[C6H5NH]
The Kb value for aniline is 4.2 x 10^-10. Let's assume that x is the concentration of hydroxide ions formed by the ionization of aniline. Then, [OH-] = x, [C6H5NH+] = x and [C6H5NH] = 0.12 - x.
Using the equation for Kb and substituting the values:
4.2 x 10^-10 = x * x / (0.12 - x)
Now, we solve for x:
x^2 = 4.2 x 10^-10 * (0.12 - x)
x^2 = 5.04 x 10^-11 - 4.2 x 10^-10 * x
Rearranging the equation:
x^2 + 4.2 x 10^-10 * x - 5.04 x 10^-11 = 0
Solving the quadratic equation using the quadratic formula, we get:
x ≈ 2.24 x 10^-6 M
Now that we have the concentration of hydroxide ions, [OH-], we can calculate the concentration of hydrogen ions [H+] using the relationship:
[H+] = Kw / [OH-]
Where Kw is the ionization constant of water, which is 1.0 x 10^-14 at 25°C. Substituting the values:
[H+] = (1.0 x 10^-14) / (2.24 x 10^-6)
[H+] ≈ 4.46 x 10^-9 M
Finally, we can calculate the pH using the equation:
pH = -log[H+]
pH = -log(4.46 x 10^-9)
pH ≈ 8.35
Therefore, the [OH-] of the 0.12 M aniline solution is approximately 2.24 x 10^-6 M, the [H+] is approximately 4.46 x 10^-9 M, and the pH is approximately 8.35.