Question

Calculate the pH of a solution prepared by mixing equal volumes of 0.20 M methylamine (CH3NH2 Kb 3.7x 10-4) and 0.60 M CH3NH3Cl.

Answer 1

Calculate the pH of a solution prepared by mixing equal volumes of 0.20 M methylamine (CH3NH2 Kb 3.7x 10-4) and 0.60 M CH3NH3Cl.

Given : Molarity of methylamine = 0.20 M , Volume lets say 1 L

Molarity of CH3NH3Cl= 0.60 M , volume lets assume = 1 L   (volume is assumed according to problem)

kb = 3.7 E-4

Now lets find moles of both.

Mol CH3NH2 = Molarity * volume in L = 0.20 M * 1.0 L = 0.20 mol CH3NH2

Mol CH3NH3Cl = 0.60 M * 1.0 L= 0.60 mol CH3NH3Cl

We know CHNH3Cl ionizes as

CH3NH3Cl ->CH3NH3+   + Cl-

CH3NH3+ is the conjugate acid of base NH3 and so this becomes a buffer system

Lets calculate concentration of both :

[CH3NH2] =mol CH3NH2/ Total volume in L = 0.20 mol / 2.0 L          (1.0 L acid + 1.0 L base)

= 0.1 M

[CH3NH3+] = [CH3NH3Cl] = 0.60 mol/ 2.0 L = 0.3 M

([CH3NH3+] = [CH3NH3Cl] ; salt dissociates completely )

Lets use Henderson equation

pOH = pkb + log ([Conjugate acid]/[conjugate base])

pkb = -log kb

Lets plug all values

= -log (3.7E-4)+ log (0.3 / 0.1)

=3.91

Now pH = 14-pOH = 14- 3.91 = 10.1