Question

A) Calculate the pH of a .01M KOH solution in water at 25 degrees C

B) What is the percentage of .01M acetic acid dissociated at 25 degrees C?

Answer 1

A. KOH is strong base . It is completely ionisable.

   KOH ---------------> K^+ + OH^-

    0.01M                                0.01M

[OH^-]     = [KOH]

[OH^-]   = 0.01M

POH = -log[OH^-]

         = -log0.01

         = 2

PH   = 14-POH

        = 14-2

         = 12

B.

       CH3COOH ---------------> CH3COO^-   + H^+

I      C                                          0                   0

C    -C                                     + C             + C

E    C-C                                   C                  C

                  Ka = C *C /C-C

                         = C^2/(1- )                   1>>>>>>>

                    Ka   = C^2

                          = Ka/C

                             = 1.8*10^-5/0.01

                             =1.8*10^-3

                             = 0.0424

                  %    = 0.0424 *100

                            = 4.24%