Question

Consider a 0.5 M  solution of   trimethylammonium chloride. The ion   trimethylammonium is the conjugate acid of the the weak base trimethylamine.The pKb  of trimethylamine is 4.2

A) Calculate [H+]

B) calculate the concentration of trimethylammonium ions

C) calculate the concentration of trimethylamine in equilibrium

D) calculate [OH-]

E) Calculate [Cl-]

F) Calculate the pH

Answer 1

Pkb = 4.2

Pka = 14-Pkb

       = 14-4.2   = 9.8

Pka = 9.8

-logKa = 9.8

      Ka = 10^-9.8   = 1.58*10^-10

                   (CH3)3NH⁺ + H2O ------> (CH3)3N + H3O⁺

        I           0.5M                                 0                 0

       C           -x                                    +x                 +x

       E         0.5-x                                 +x                +x

                        Ka =   [(CH3)3N] [H3O⁺ ]/[(CH3)3NH⁺ ]

                        1.58*10^-10   = x*x/0.5-x

                         1.58*10^-10 *(0.5-x) = x²

                       x    = 8.8*10^-6

                 [H3O+] = x = 8.8*10^-6 M

                  pH   = -log[H3O+]

                          = -log8.8*10^-6

                           = 5.0555

              [OH-] = Kw/[H3O+]

                        = 1*10^-14/8.8*10^-6

                        = 1.13*10^-9 M

    [Cl-]   = 0.5M

B) the concentration of trimethylammonium ions   = 0.5M

C) the concentration of trimethylamine in equilibrium = 0.5-x = 0.5-0.0000088 = 0.4999M