Question

1 Reviewid Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions. Part A 0.15 M CH3NH2 (where K for CH3NH2 is 4.4 x 10-4) Express your answer using two decimal places. O A A O O ? pH = Submit Request Answer Part B 0.15 M CH3NH2C1 Express your answer using two decimal places. IV AEQ * R 0 2 ? pH = Submit Request Answer

Answer 1

Answer A:

Kb = 4.4 x 10-4

Acid-base reaction:

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

Initial 0.15 0 0

Equilibrium 0.15 - x x x

So, Kb = [CH3NH3+] [OH-]/ [CH3NH2]

Kb = (x) (x)/ (0.15-x)

4.4 x 10-4 = x2/(0.15-x)

Here, (0.15−x)≅0.15.

So, 4.4 x 10-4 = x2/0.15

x2 = 0.66 x 10-4

x = 0.812 x 10-2 = 0.00812 = [OH-]

pOH = -log [OH-] = - log 0.00812 = -(-2.09) = 2.09

pH+pOH = 14

pH = 14-pOH = 14-2.09 = 11.91