Question

50. mL 2.0 M methylamine (CH₃NH₂) solution and 50. mL of 2.0 M methylamine hydrochloride (CH₃NH₃Cl) solution. The K_b for methylamine is 4.4 x 10^-4.

What is the pH of the resulting solution after adding 100 mL of 1.0 M KOH to the original buffer solution?

11.12                           b) 10.64                       c) 12.32                       d) 0.48                         e) 5.32

Answer 1

mmillimoles of CH3NH2 = 50 x 2 = 100

millimoles of CH3NH3Cl = 50 x 2 = 100

millimolesof KOH added = C = 100 x 1 = 100

KOH is base and CH3NH3Cl is so acid completely react with base . buffer will distoyed .

now only CH3NH2 remains

CH3NH2 concentration = 100 / (50 +50 +100) = 0.5 M

CH3NH2 + H2O ------------------------> CH3NH3+ + OH-

0.5 -x                                                       x                  x

Kb = x^2 / 0.5-x

4.4 x 10^-4 = x^2 / 0.5-x

x^2 + 4.4 x 10^-4 x - 2.2 x 10^-4 = 0

x = 0.0146

[OH-] = 0.0146 M

pOH = -log [OH-]

pOH = 1.84

pH + POH = 14

pH = 12.3

answer : C) 12.32