2. For the diprotic weak acid H2A, Ka1 = 2.7 × 10-6 and Ka2 = 8.8...

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Chemistry

2. For the diprotic weak acid H2A, Ka1 = 2.7 × 10-6 and Ka2 =8.8 × 10-9. What is the pH of a 0.0500 M solution of H2A? What arethe equilibrium concentrations of H2A and A2– in this solution?

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3.9 Ratings (617 Votes)

             H2A ---------> HA-   + H+

   I          0.05                0          0

C         -x                    +x        +x

   E       0.05-x               +x        +x

         ka1   = [H+][HA-]/[H2A]

         2.7*10-6   = x*x/0.05-x

         2.7*10-6 *(0.05-x) = x2

             x = 0.000366

         [H+] = x = 0.000366M

         [HA-]    = x = 0.000366M

        HA- -------> A2- + H+

   Ka2 = [A2-][H+]/[HA-]

8.8*10-9   = [A2-]*0.000366/0.000366

   [A2-]   = 8.8*10-9 M

PH = -log[H+]

         = -log0.000366   = 3.4365


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