Question

Calculate the concentration (in M) of each species present in a 0.085 M solution of H2GeO3. (See the Acid and Base Dissociation Constants table. Assume Kw = 1.01 ✕ 10−14.)

[H2GeO3]______ M

[HGeO3− ]______ M

[GeO32− ]______ M

M [H3O+ ]______ M

M [OH − ]_______ M

Answer 1

ka1   = 9.7*10^-10

Ka2   = 5*10^-13

          H2GeO3 -------------> H^+ (aq) + HGeO3^-

I         0.085                         0                   0

C         -x                              +x               +x

E       0.085-x                        +x              +x

          Ka   = [H^+][HGeO3^-]/[H2GeO3]

         9.7*10^-10 = x*x/0.085-x

       9.7*10^-10 *(0.085-x) = x^2

         x = 9.08*10^-6

        [H^+]   = x = 9.08*10^-6M

        [HGeO3^-] = x = 9.08*10^-6M

         [H2GeO3]   = 0.085-x = 0.085-9.08*10^-6   = 0.085M

   [OH^-]   = Kw/[H^+]

                 = 1*10^-14/9.08*10^-6   = 1.1*10^-9M

    HGeO3 ^- ------------> GeO3^2-   + H^+

     Ka2    = [GeO3^2-][H^+]/[HGeO3^-]

    5*10^-13   = [GeO3^2-] *9.08*10^-6/9.08*10^-6

[GeO3^2-]   = 5*10^-13M