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 K_b for methylamine is 4.4 x 10^-4.

A. What is the pH of the resulting solution after adding 200 ml of 1.0 M HCl to the original buffer solution?

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

B.What is the pH of the resulting solution after adding 50 mL of 1.0 M KOH to the original buffer solution?

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

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

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

Answer 1

A) option d) 0.48    
millimoles of CH₃NH_{2 =} 50 x 2 = 100

millimoles of CH₃NH₃Cl = 50 x 2 = 100

millimoles of HCl = 200 x 1 = 200

millimoles of HCl > base

strong acid reamins = 200 - 100 / (200 + 50 + 50)

                                = 0.33 M

pH = -log [H+] = -log 0.33

pH = 0.48

B) option a) 11.12

millimoles of KOH = 50 x 0.1 = 50

pOH = pKb + log [100 - 50 / 100 + 50]

       = 3.36 + log [50 / 150]

     = 2.88

pH = 11.12

C.) option C)