Question

Calculate the pH and concentrations of CH3NH2 and CH3NH3 in a 0.0367 M methylamine (CH3NH2) solution. The Kb of CH3NH2 = 4.47 × 10-4.

pH=

[CH3NH2]=

[CH3NH3+]=

Answer 1

pH = 11.58

[CH3NH2] = 0.03287M

[CH3NH3⁺] = 0.003833M

Explanation

The ionization equillibrium of CH3NH2 is

CH3NH2(aq) + H2O(l) <------> CH3NH3⁺(aq) + OH^-(aq)

K_b = [CH3NH3⁺] [OH^-]/[CH3NH2] = 4.47 ×10^-4

Initial concentration

[CH3NH2] = 0.0367

[CH3NH3⁺] = 0

[OH^-] = 0

change in concentration

[CH3NH2] = -x

[CH3NH3⁺] = +x

[OH^-] = +x

Equillibrium concentration

[CH3NH2] = 0.0367 - x

[CH3NH3⁺] = x

[OH^-] = x

so,

x²/(0.0367 - x) = 4.47 ×10^-4

solving for x

x = 0.003833

Therefore, at equillibrium

[CH3NH2] = 0.0367 - 0.003833 = 0.03287M

[CH3NH3⁺] = 0.003833M

[OH^-] = 0.003833M

pOH = -log[OH^-]

pOH = -log(0.003833) = 2.42

pH = 14 - pOH

pH = 14 - 2.42

pH = 11.58