Question

What is the Kb of a 0.0200 M (at equilibrium) solution of methylamine, CH3NH2, that has a pH = 11.40?

A.8.61


B. 7.92 × 10−22


C. 1.26 × 10−7


D. 3.15 × 10−4

Answer 1

Answer: C.

Explanation: It's what I put and I got it correct.

Answer 2

The answer is D. 3.15 × 10−4

Now the dissolution of methylamine can be represented as follows:

CH3NH2 + H2O → CH3NH3OH

The product formed dissociates and it is represented as given below

CH3NH3OH ↔ CH3CH3+ + OH-

Now pH = 11.4 ,

Then pOH = 2.6

[OH-] = 10^-2.6

[OH-] = 2.51*10^-3

Considering the below equation again

CH3NH3OH ↔ CH3CH3+ + OH-

We can calculate Kb = [CH3NH3] [ OH-] / [CH3NH3OH]

where Kb is the equilibrium constant.

Now [CH3NH3] = [OH-] = 2.51 × 10^-3

and [CH3NH3OH] = 0.020M (given)

Substituting these values we get

Kb = ( 2.51 × 10^-3)² / 0.02

Kb = 6.31 × 10^-6 / 0.02

Kb = 3.15 × 10^-4