Question

Calculate the pH and the equilibrium concentrationof the CO32-ion of an aqueous 0.050 M carbonic acid (H2CO3) solution at 25 oC. The acid dissociation constants for carbonic acid are Ka1= 4.3 × 10-7, and Ka2= 5.6× 10-11respectively.

Answer 1

             H2CO3 ---------------> H^+ + HCO3^-

I          0.05                           0            0

C            -x                             +x             +x

E         0.05-x                         +x           +x

         Ka1    =   [H+][HCO3^-]/[H2CO3]

          4.3*10^-7    =   x*x/(0.05-x)

          4.3*10^-7 *(0.05-x)   = x^2

           x   = 0.000146

          [H+}   = x   = 0.000146M

         PH   = -log[H+]

                  = -log0.000146

                   = 3.8356

HCO3 ^- ---------------> H^+ + CO3^2-

     Ka2   = [H^+][CO3^2-]/[HCO3^-]

     5.6*10^-11     = 0.000146*[CO3^2-]/0.000146

[CO3^2-]         = 5.6*10^-11 M