Question

What is the pH of a 0.136 M monoprotic acid whose Ka is 5.7 10-4?

Answer 1

                  HA --------> H⁺ + A^-

I              0.136          0        0

C              -x               +x      +x

E              0.136-x         +x      +x

         Ka = [H⁺][A^-]/[HA]

        5.7*10^-4   =   x*x/0.136-x

        5.7*10^-4 *(0.136-x) = x²

           x   = 0.0085

         [H+]   = x = 0.0085M

       PH    = -log[H+]

                = -log0.0085

                = 2.0705