Question

A solution of aniline (C6H5NH2, Kb = 4.2 x 10-10) has a pH of 8.79 at 25°C. What was the initial concentration of aniline?

a) 6.2 x 10⁻⁶ M
b) 7.8 x 10⁻¹M
c) 1.6 x 10⁻¹M
d) 9.1 x 10⁻² M
e) 4.2 x 10⁻¹⁰M

Answer 1

To find the initial concentration of aniline when the pH is 8.79, the pOH is first determined and then used to find the hydroxide [OH-] concentration. This concentration is used in the equilibrium expression for aniline to solve for the initial concentration, which is approximately 6.2 x 10^-6 M.

Step-by-step explanation:

The question involves calculating the initial concentration of aniline in a solution when the pH is given. The pH of the solution is 8.79, and the Kb for aniline is 4.2 x 10-10. First, we need to find the concentration of OH- by using the pH and the relation pOH = 14 - pH. This gives us a pOH of 5.21, and thus a concentration of OH- of 10-pOH = 6.2 x 10-6 M.

The equilibrium expression for aniline in water is Kb = [C6H5NH3+][OH-]/[C6H5NH2]. We can let the initial concentration of aniline be x and the concentration of OH- which comes from the ionization of aniline also be x. Thus, Kb = x2/x and since Kb = 4.2 x 10-10, we can solve for x and find that the initial concentration of aniline is approximately 6.2 x 10-6 M.