Question

THis is a ph Equilibirum problem.

10^(-2)M of acetic acid at 25 degrees Celcius. What are the ion concentrations?

ka= 1.8x10^-5

Final Answers:

[H+]= 4.15 x10 ^-4

[HAc]=9.59x10^-3

[OH+]=2.41x10^-11

[Ac-]=4.15x10^-4

Answer 1

       CH3COOH ------------------> CH3COO^-    + H^+

       HAC -------------------> AC^-   + H^+

I                   0.01                           0               0

C                    -x                            +x              +x

E                  0.01-x                      +x                +x

           Ka   = [AC^-][H^+]/HAC]

           1.8*10^-5   = x*x/(0.01-x)

           1.8*10^-5 *(0.01-x) = x^2

               c   = 0.000415

      [AC^-]    = x   = 0.000415M = 4.15*10^-4M

      [H^+]    = x     = 0.000415M = 4.15*10^-4M

      [HAC]   = 0.01 -x   = 0.01-0.000415   = 0.00959M = 9.59*10^-3M

       [OH^-]    = Kw/[H^+]

                      = 1*10^-14/4.15*10^-4   = 2.41*10^-11 M