Question

The following buffer was prepared: 50. mL 2.0 M methylamine (CH3NH2) solution and 50. mL of 2.0 M methylamine hydrochloride (CH3NH3Cl) solution. The Kb for methylamine is 4.4 x 10-4

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

A) 11.12 B) 10.64 C) 12.32 D) 0.48 E) 5.32 (Correct Answer: 5.32) Please show work, and explain why.

Answer 1

Set up a reaction table for the following ionization:

                      CH₃NH₂ + H₂O ↔ CH₃NH₃⁺ + OH^-

Initial            2.0                        0             0

Equilibrium         2.0-x                      x             x

K_b = [CH₃NH₃⁺][OH^-]/[CH₃NH₂]

4.4 10^-4 = x²/2.0

x = 0.0297 = [OH^-]

pOH = -log[OH^-] = -log(0.0297) = 1.53

pH = 14-pOH = 14-1.53 = 12.47

pH = pK_a + log ([A^-]/[HA])

12.47 = pK_a + log (2.0/2.0)

pK_a = 12.47 = -log[K_a]

K_a = 3.39 10^-13

               CH₃NH₂ + H⁺ ↔ CH₃NH₃⁺

Initial           2.0       1.0         1.0     

Equili         2.0-x    1.0+x      1.0+x

K_a = [CH₃NH₃⁺]/[CH₃NH₂][H⁺]

3.39 10^-13 = (1.0+x)²/(2.0-x)

x = 4.78 10^-6 = [H⁺]

pH = -log[H⁺] = -log[4.78 10^-6] = 5.32