Question

Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH₃NH₂ and 0.10 M in C₅H₅N K_b=1.7x10^-9

Answer 1

The general form of an acid base equation that has Kb for an equilibrium constant has the base + water on the left and the acid + OH^- on the right. Methylamine is the base and the methylammonium ion is the acid. So the acid-base equation can be written in this way:

CH₃NH₂ + H₂O <---> CH₃NH₃⁺ + OH^-

I 0.25M 0 0 0

C   -x +x +x

E   0.25M - x x x

The equilibrium constant of this equation is the Kb of methylamine, so Kb:

Kb = 4.4x10^-4 = [CH₃NH₃⁺][OH^-] [CH₃NH₂]

For the acid base equation for Kb, we leave water out of the equation.

4.4x10^-4 = x² 0.25 - x = x² 0.25 because 0.25 -x = 0.25

Solving for x, we get:

x² = 0.00044 0.25 = 0.00176

x² = 0.00176 = 0.042

Therefore, x = [OH] = 0.042

We know that pOH = - log [OH] = - (- 1.38) = 1.38

pH = 14 - pOH = 14 - 1.38 = 12.62

The general form of an acid base equation that has Kb for an equilibrium constant has the base + water on the left and the acid + OH^- on the right. Pyridine is the base and the pyridinium ion is the acid. So the acid-base equation can be written in this way:

C₅H₅N + H₂O <---> C₅H₅NH⁺ + OH^-

I 0.10M 0 0 0

C   -x +x +x

E   0.10M - x x x

The equilibrium constant of this equation is the Kb of pyridine, so Kb:

Kb = 1.7 10^-9 = [C₅H₅NH⁺][OH^-] [C₅H₅N]

For the acid base equation for Kb, we leave water out of the equation.

1.7 10^-9 = x² 0.10 - x = x² 0.10 because 0.10 -x = 0.10

Solving for x, we get:

x² = 0.0000000017 0.10 = 0.000000017

x² = 0.000000017 = 0.0001304

Therefore, x = [OH] = 0.0001304

We know that pOH = - log [OH] = - ( - 3.885) = 3.885

pH = 14 - pOH = 14 - 3.885 = 10.12