Chemistry Word Problem

Question

Calculate the pH and concentrations of ch3nh2 and ch3nh3+ in a 0.0349 M methylamine (ch3nh2) solution. the Kb of ch3nh2 is 4. 47×10−4.

Answer 1

To answer this question, we will first look at the hydrolysis of methylamine and using the Kb, we will find the [OH^-], since methylamine is a base. Once we have the [OH^-], we can easily find the pOH and then the pH.

CH₃NH₂ + H₂O ==> CH₃NH₃⁺ + OH^- .. hydrolysis of methylamine

Kb = [CH₃NH₃⁺] [OH^-] / [CH₃NH₂]

4.47x10^-4 = (x)(x) / 0.0349 - x

x² = 1.54x10^-5 - 4.47x10^-4x

x² + 4.47x10^-4x - 1.54x10^-5 = 0

Solving for x using the quadratic formula we get ...

x = [OH^-] = 0.00371 M

pOH = -log [OH^-] = -log 0.00371

pOH = 2.43

pH = 14 - pOH = 14 - 2.43

pH = 11.57

Added: Forgot to calculate [CH₃NH₂] and [CH₃NH₃⁺] ...

[CH₃NH₂] = 0.0349 M - 0.00371 M = 0.0312 M (3 sig. figs)

[CH₃NH₃⁺] = 0.00371 M (3 sig. figs.)