Question

What is the kb of a 0.0200 m ( at equilibrium) solution of methyl amine, ch3nh2, that has a ph of 11.40?

Answer 1

Answer: Kb = 3.15 × 10 ⁻⁴

Explanation:


This is how you calculate Kb for this reaction.

1) Equilibrium equation:

CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻

2) Kb = [CH₃NH₃⁺]  [OH⁻] / [CH₃NH₂] ↔ all the spieces in equilibrium


3) From the stoichiometry [CH₃NH₃⁺] = [OH⁻]

Then, Kb = [OH⁻]  [OH⁻] / [CH₃NH₂] = [OH⁻]² / [CH₃NH₂]

4) You get [OH⁻] from the pH in this way:

pOH + pOH = 14 ⇒ pOH = 14 - pH = 14 - 11.40 = 2.60

pOH = - log [OH⁻] = 2.60 ⇒ [OH⁻] = 10^(-2.6) = 0.002512

5) [CH₃NH₂] in equilibrium is given: 0.0200M


6) Now compute:

Kb = (0.002512)² / 0.0200 = 3.15 × 10 ⁻⁴