Question

a. A 0.037 M solution of a weak acid is 4.67 percent ionized in solution. What is the Ka for this acid?

b. The pOH of a 0.29 M weak acid solution is 9.7. What is the Ka for this weak acid?

Answer 1

               HA -------------> H^+ (aq) + A^-

I             C                       0                0

C           -C                 C               C

E           C-C               C                C

              C(1-)

                  Ka = [H^+][A^_]/[HA]

                Ka   = C*C/C(1-)

               Ka    = C^2                                       1>>>

                     = Ka/C

               0.0467   = Ka/0.037

              squring on both sides

             0.00218     = Ka/0.037

              Ka             = 0.00218*0.037   = 8.066*10^-5

b.   PH = 14-POH

              = 14-9.7 = 4.3

        PH   = 1/2Pka -1/2logc

      4.3     = 1/2Pka -1/2log0.29

     4.3     = 1/2Pka -1/2*-0.5376

    4.3    = 1/2Pka+0.2688

   1/2Pka   = 4.3-0.2688

   1/2Pka    = 4.0312

       Pka    = 8.0624

      -logka = 8.0624

         Ka      = 10^-8.0624   = 8.67*10^-9 >>>answer