To find the pka of x-281, you prepare a 0.089 m test solution of x-281 at 25.0 ∘c. the ph of the solution is determined to be 3.00. what is the pka of x-281?

pKa = 4.96

If X-281 is a weak acid HA, it will hydrolyze in water to produce some H3O+ and an equal amount of A-.  

                            HA + H2O ==> H3O+ + A-  

Initial molarity ;    0.095 . . . . . . . . . .0 . . . .0  

Change;                   -x . . . . . . . . . . . .x . . . .x  

At Equilibrium;   0.095 - x . . . . . . . . .x . . . .x  

Ka = [H3O+][A-] / [HA]

      = x^2 / (0.095 - x)  

But since the pH = 3.00, then [H3O+]

                            = 10^-pH = 1 x 10^-3 M

                            = x = 0.001.  

Ka = (0.001)^2 / (0.095 - 0.001)

     = 1.1 x 10^-5

But; pKa = -log (Ka)  

       pKa = - log (1.1 x 10^-5)

              = 4.96

i'm pretty sure its b i might be wrong though!