Calculate the pH and concentrations of CH3NH2 and CH3NH+3 in a 0.0319 M methylamine (CH3NH2) solution. The 𝐾b of CH3NH2 is 4.47×10−4.

Answer 1

CH3NH2 is a weak base

CH3NH3+ is the conjugate acid of that weak base

Together, these form a buffer.

The Henderson Hasselbach equation can generally be used to calculate pH of a buffer. This equation is...

pOH = pKb + [conj.acid] /[ base]

Looking at the hydrolysis of CH3NH2, we have...

CH3NH2 + H2O ==> CH3NH3⁺ + OH^-

Kb = 4.47x10^-4 = [CH3NH3⁺][OH^-] / [CH3NH2]

4.47x10^-4 = (x)(x) / 0.0319-x)

4.47x10^-4 = x² / 0.0319-x

x² = 1.43x10^-5 - 4.47x10^-4x

x² + 4.47x10^-4x - 1.43x10^-5 = 0

x = [CH3NH3⁺] = [OH^-] = 0.00356 M

So, [CH3NH2] = 0.0319 M - 0.00356 M = 0.0283 M

[CH3NH3⁺] = 0.00356 M

pH = 14 - pOH

pOH = -log [OH-] = -log 0.00356

pOH = 2.45

pH = 14 - 2.45

pH = 11.6

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Calculation using the Henderson Hasselbalch equation:

pOH = pKb + log [CH3NH3⁺] / [CH3NH2] and pKb = -log Kb = -log 4.47x10^-4 = 3.35

pOH = 3.35 + log (0.00356/0.0283)

pOH = 3.35 + log 0.126

pOH = 3.35 -0.9-0

pOH = 2.45

pH = 11.6 (agrees with above calculation)