Calculate [OH − ], [H+], and the pH of a 0.12 M solution of Aniline. The Kb is 3.8 e-10
To calculate [OH-], [H+], and the pH of a solution of aniline, we need to use the Kb value.
Kb is the equilibrium constant for the reaction of aniline (C6H5NH2) with water to form the anilinium ion (C6H5NH3+) and hydroxide ion (OH-):
C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-
The Kb expression for this reaction is:
Kb = [C6H5NH3+][OH-] / [C6H5NH2]
Given that the Kb value is 3.8e-10, we can set up the following equation:
3.8e-10 = [C6H5NH3+][OH-] / [C6H5NH2]
Since we are given a concentration of 0.12 M for aniline (C6H5NH2), we can substitute this value into the equation:
3.8e-10 = [C6H5NH3+][OH-] / 0.12
Now we can solve for [OH-]:
[OH-] = (3.8e-10)(0.12) / [C6H5NH3+]
However, to determine the concentration of [C6H5NH3+], we need to know the initial concentration of aniline and the extent of its ionization.
Without more information, we cannot calculate [OH-], [H+], or the pH of the solution.