Calculate the concentrations of all species present C6H5N2, C6H5NH3+ OH^-, and H3O^+}) in a 0.14 M solution of aniline.
To calculate the concentrations of all species present in a solution of aniline (C6H5NH2), we need to consider the acid-base equilibrium of aniline.
Aniline (C6H5NH2) can act as both a weak base and a weak acid. In water, it can accept a proton (H+) to form the anilinium ion (C6H5NH3+), or it can donate a proton to form the hydroxide ion (OH^-).
First, we need to write the balanced equation for the acid-base reaction of aniline:
C6H5NH2 + H2O ⇌ C6H5NH3+ + OH^-
Next, we can define the equilibrium constant for this reaction, known as the base ionization constant (Kb), using the concentrations of the species involved:
Kb = [C6H5NH3+][OH^-] / [C6H5NH2]
We are given that the solution has a concentration of aniline (C6H5NH2) of 0.14 M. Therefore, the concentration of the anilinium ion ([C6H5NH3+]) and the hydroxide ion ([OH^-]) will initially be zero.
Let's assume x moles of aniline (C6H5NH2) react with water (H2O) to form x moles of anilinium ion (C6H5NH3+) and x moles of hydroxide ion (OH^-).
At equilibrium, the concentration of aniline (C6H5NH2) will be (0.14 - x) M, the concentration of anilinium ion (C6H5NH3+) will be x M, and the concentration of the hydroxide ion (OH^-) will also be x M.
Using the equilibrium constant expression, we can write:
Kb = [C6H5NH3+][OH^-] / [C6H5NH2]
Kb = (x)(x) / (0.14 - x)
Now, we can solve for x by substituting the given Kb value and solving the resulting quadratic equation.
Finally, once we know the value of x, we can determine the concentrations of all species present:
[C6H5NH2] = 0.14 - x (M)
[C6H5NH3+] = x (M)
[OH^-] = x (M)
[H3O^+] = 10^(-pH)
Note: The concentration of [H3O^+] can be determined using the pH of the solution, and it is not directly related to the acid-base equilibrium of aniline.