Question

Calculate the pH and [H3O+] of the following solutions:

a) 0.20 M potassium hydroxide (KOH, strong base)

b) 0.10 M propanoic acid (CH3CH2COOH, Ka = 1.3 x 10-5)

c) 1.50 M diethyl amine ((CH3)2NH, Kb = 5.9 x 10-4)

Answer 1

a. KOH ---------> K⁺ + OH^-

     0.2M                      0.2M

[OH-]   = [KOH]

[OH-] = 0.2M

[H3O+] = Kw/[OH-]

               = 1*10^-14/0.2

               = 5*10^-14 M

PH   = -log[H3O+]

       = -log5*10^-14

      = 13.3010

b.                 CH3CH2COOH + H2O ---------> CH3CH2COO- + H3O⁺

          I            0.1                                                 0                     0

         C           -x                                                   +x                   +x

         E          0.1-x                                                +x                 +x

                  Ka = [CH3CH2COO-][H3O+]/[CH3CH2COOH]

                   1.3*10^-5    = x*x/0.1-x

                   1.3*10^-5 *(0.1-x) = x²

                         x   = 0.001133

                      [H3O+] = x = 0.001133M

                    PH   = -log[H3O+]

                            = -log0.001133

                            = 2.9457

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

       I         1.5                                         0                      0

     C          -x                                          +x                     +x

     E        1.5-x                                       +x                     +x

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

            5.9*10^-4    = x*x/1.5-x

            5.9*10^-4 *(1.5-x) = x²

                 x = 0.02945

             [OH^-] = x = 0.02945M

            [H3O+] = Kw/[OH^-]

                          = 1*10^-14/0.02945

                          = 3.395*10^-13 M

           PH        = -log[H3O+]

                       = -log3.395*10^-13

                      = 12.4691