Question

In the solution containing both 0.10 M acetic acid and 0.10 M sodium acetate, the acetic acid undergoes ionization. The chemical equation for this ionization reaction is the same as for a solution containing acetic acid alone. The difference is that the initial concentration of acetate ion (before any ionization reaction occurs) for the solution containing acetic acid alone is zero, whereas the initial concentration of acetate ion is 0.10 M in your solution containing both acetic acid and sodium acetate. Calculate the percent ionization and the expected initial pH for the solution that contained both 0.10 M acetic acid and 0.10 M sodium acetate. (Hint: Again, you will need to use Ka for acetic acid.) Percent Ionization Calculated pH

Answer 1

pka of aceticacid = -logka

             = -log(1.75*10^−5)
           
             = 4.76

aceticacid + CH3COONa = buffer

pH of acidic buffer = pka + log(salt/acid)

                      = 4.76 + log(0.1/0.1)

                      = 4.76

             pH = 4.76

           CH3COOH (aq) <-----> CH3COO-(aq) + H^+(aq)

   initial 0.1 M                  0.1 M       0

   change    x                       x         x

equilibrium 0.1-x               0.1+x        x

    ka = [CH3COO-][H+]/[CH3COOH]

    1.75*10^-5 = ((0.1+X)X)/(0.1-X)

    X = 1.75*10^-5 M

percentage dissociation of CH3COOH = dissociated / initial*100

                = (1.75*10^-5)/0.1*100

               = 0.0175 %