Question

The Common Ion Effect: Calculation of pH. When a weak, nitrogen base is protonated, the resulting conjugate acid is given the "ium" suffix to show that it has been protonated and has an overall +1 charge. Thus, the weak base ammonia is protonated to form the ammonium (NH4+) cation. What is the pH of a 0.68 M solution of (CH3)2NH that is also 0.65 M in dimethylammonium sulfate, the salt of it's conjugate acid? The Kb for (CH3)2NH = 5.4 X 10-4. Is the pH of the this solution higher, lower or the same as the pH for just a 0.68 M solution of (CH3)2NH higher lower same

Answer 1

The Kb for (CH3)2NH = 5.4 X 10-4.

PKb   = -logKb

         = -log5.4*10^-4

         = 3.2676

POH   = PKb + log[(CH3)2NH2SO4]/[(CH3)2NH]

          = 3.2676 + log0.65/0.68

            = 3.2676 -0.01959

           = 3.248

PH   = 14-POH

         = 14-3.248

        = 10.752

          (CH3)2NH(aq) + H2O(l) ----------------> (CH3)2NH2^+ (aq) + OH^- (aq)

I         0.68                                                                0                          0

C          -x                                                                  +x                         +x

E        0.68-x                                                              +x                          +x

        kb    = [(CH3)2NH2^+][OH^-]/[(CH3)2NH]

        5.4*10^-4   = x*x/0.68-x

       5.4*10^-4 *(0.68-x) = x^2

               x = 0.0189

   [OH^-]   =x    = 0.0189M

POH = -log[OH^-]

            = -log0.0189

               = 1.7235

PH     = 14-POH

          = 14-1.7235

            = 12.2765

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