Question

Calculate the pH of a 0.0333 M aqueous solution of dimethylamine ((CH₃)₂NH, K_b = 5.9×10^-4) and the equilibrium concentrations of the weak base and its conjugate acid.

pH	=
[(CH3)2NH]equilibrium	=	M
[(CH3)2NH2+ ]equilibrium	=	M

Answer 1

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

     I             0.0333                               0                     0

    C             -x                                      +x                     +x

   E            0.0333-x                               +x                     +x

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

       5.9*10^-4     = x*x/0.0333-x

      5.9*10^-4 *(0.0333-x) = x^2

       x= 0.004147

     [OH-] = x = 0.004147M

[(CH₃)₂NH] = 0.0333-x     = 0.0333-0.004147 = 0.0292M

[(CH₃)₂NH₂⁺ ]   = x= 0.004147M

   POH   = -log[OH-]

             = -log0.004147   = 2.3823

PH    = 14-POH

           = 14-2.3823 = 11.6177