Question

What is the pH of a 0.402 M aqueous solution of NaCH3COO? Ka (CH3COOH) = 1.8x10-5

Answer 1

                    CH3COONa ----------> CH3COO- + Na⁺

                0.402M                             0.402M

             CH3COO^-   + H2O --------> CH3COOH + OH^-

    I         0.402M                                 0                0

C          -x                                         +x               +x

   E        0.402-x                                 +x                +x

           kb   = Kw/Ka

                  = 1*10^-14/1.8*10^-5   = 5.6*10^-10

     Kb   = [CH3COOH][OH-]/[CH3COOH]

   5.6*10^-10   = x*x/0.402-x

5.6*10^-10 *(0.402-x) = x²

     x = 1.5*10^-5

   [OH-] = x=1.5*10^-5 M

POH = -log[OH-]

             = -log1.5*10^-5

            = 4.8239

PH   = 14-POH

         = 14-4.8239   = 9.1761