Question

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

Why won't it let me answer this? I have the correct answer and a detailed explanation, but the answer on there currently is incorrect. they find an incorrect Ka because they use the concentration of the base in the denominator of the Ka equation, which should be the acid.

You get the pOH from 14-11.4=2.6 and then get OH- concentration= 10^-2.6 and the Kb= (10^-2.6)^2/.02M so Kb= 3.15 × 10−4

Answer 1

Answer: Kb = 3.15 × 10 ⁻⁴

Step-by-step 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