Calculate the pH and concentrations of CH3NH2 and CH3NH+3 in a 0.0459 M methylamine (CH3NH2) solution.

Question

Calculate the pH and concentrations of CH3NH2 and CH3NH+3 in a 0.0459 M methylamine (CH3NH2) solution. The Kb of CH3NH2 is 4.47×10^−4.

Answer 1

First, I assume CH3NH+3 means CH3NH3+, that is the protonated CH3NH2.

CH3NH2 + H2O ----> CH3NH3+ + OH-

The ionization equilibrium is Kb = [CH3NH3+] x [OH1-] / {[CH3NH2] - [OH-]}

As one OH- is produced for every CH3NH3+, we can write Kb = [OH-] ^2 / {[CH3NH2] - [OH-]}.

Rearrange to get a quadratic equation [OH-]^2 +Kb x [OH-] - 0.0459 x Kb = 0

Solve for [OH-] using the quadratic formula, discarding the negative root.

[OH-] = -Kb ± √ Kb^2 -4 x 1 x (-0.0459 x Kb) /2 = 8.6 10^-3

[H+] = 10^-14 / 8.6 x 10^-3 = 1.16 x 10^-12

pH = - log (1.16 x10^-12) = 11.9

I did not make the assumption that the degree of dissociation is negligible.

Check the math.