Question

Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions.

a) 0.13 mol*L−1 CH3NH2

b) 0.13 mol*L−1 CH3NH3Cl

c)a mixture that is 0.13 molL−1 in CH3NH2
and 0.13 molL−1 in CH3NH3Cl

Answer 1

pH calculation

a) 0.13 M CH3NH2

ICE chart

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

I               0.13                                      -                    -

C               -x                                       +x                 +x

E           0.13 - x                                    x                  x

So,

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

1.8 x 10^-5 = x^2/(0.13 - x)

x^2 + 1.8 x 10^-5 - 2.34 x 10^-6 = 0

x = [OH-] = 1.52 x 10^-3 M

pOH = -log[OH-] = 2.82

pH = 14 - pOH = 11.18

b) 0.13 M CH3NH3Cl

             CH3NH3+    + H2O <===> CH3NH2    +   H3O+

I               0.13                                      -                    -

C               -x                                       +x                 +x

E           0.13 - x                                    x                  x

So,

Ka = Kw/Kb = [CH3NH2][H3O+]/[CH3NH3+]

1 x 10^-14/1.8 x 10^-5 = x^2/(0.13 - x)

x^2 + 5.55 x 10^-10x - 7.22 x 10^-11 = 0

x = [H3O+] = 8.50 x 10^-6 M

pH = -log[H3O+] = 5.07

c) Buffer solution

pH = pKa + log(base/acid)

      = 9.25