The pH of an aniline solution is 8.91, and the solution has a concentration of 0.17 M. The pK_b of aniline is 9.42. What is the percent ionization of aniline?

Question

Quizplus · Accepted Answer

To find the percent ionization of aniline, we need to use the expression for the equilibrium constant (Ka) for the reaction:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

Ka = ([C6H5NH3+][OH-])/[C6H5NH2]

At equilibrium, [C6H5NH3+] = [OH-] (since one hydrogen ion is released for each hydroxide ion in the reaction), so we can write:

Ka = [OH-]^2/[C6H5NH2]

pKa = -log(Ka) = 9.42
pH = 8.91

Since aniline is a weak base, we can assume that [OH-] is much smaller than [C6H5NH2]. This allows us to simplify the expression for Ka to:

Ka = [OH-]^2/0.17

Solving for [OH-], we get:

[OH-] = sqrt(Ka x [C6H5NH2]) = sqrt(10^-9.42 x 0.17) = 5.84 x 10^-6 M

The percent ionization of aniline is given by:

% ionization = ([OH-]/[C6H5NH2]) x 100%
% ionization = (5.84 x 10^-6/0.17) x 100% = 0.047%, which is approximately 4.7%.

Therefore, the correct answer is D.

The PH of an Aniline Solution Is 8 ×\times× 10-3% D)4 ×\times×

The PH of an Aniline Solution Is 8 $\times$ 10^-3%
D)4 $\times$