What is the ph of a solution prepared by mixing 50.00 ml of 0.10 m methylamine, ch3nh2, with 20.00 ml of 0.10 m methylammonium chloride, ch3nh3cl? assume that the volume of the solutions are additive and that kb = 3.70 × 10-4 for methylamine

pH=10.97

Explanation:

the solution of methyl amine with methylammonium chloride will make a buffer solution.

The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:

pH = 14- pOH

Let us calculate pOH

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

[base] = [Methylamine]=0.10

After mixing with salt

pKb= -log[Kb]= 3.43

Putting values

pOH =