Question

cetylsalicylic acid, a monoprotic weak acid, is the active ingredient in aspirin. Two aspirin tablets, each containing 500.0 mg of acetylsalicylic acid, are dissolved in 25.0 mL water. a. Calculate the pH of the solution. b. Calculate the percent ionization of acetylsalicylic acid in the solution. (2 pts)

Answer 1

molarity =   W*1000/G.M.Wt*volume of solution in ml

                = 500*10^-3 *1000/180.15 *25    = 0.111M

   HA   ------------> H⁺ + A^-

I          0.111                0        0

C         -x                      +x        +x

E        0.111-x                +x       +x

      Ka   = [H+][A-]/[HA]

    3*10^-4   = x*x/0.111-x

   3*10^-4 *(0.111-x) = x²  

            x= 0.0056

[H+] = x = 0.0056M

PH = -log[H+]

        = -log0.0056   = 2.2518

[H+] = C*ionisation

ionisation% = [H+]*100/C

                       = 0.0056*100/0.111 = 5.045%