Buffers And Acid-Base Titrations

 

16) (a) mNH3 = 5.0 g NH3          mNH4Cl = 20.0 g NH4Cl    

            V = 2.50 L                      Kb(NH3) = 1.8 x 10-5

 

            NH4Cl(aq) NH4+(aq) + Cl-(aq)

 

            Kw = Ka x Kb = 1.00 x 10-14

            Ka = 1.00 x 10-14/1.8 x 10-5 = 5.6 x 10-10

 

            [NH3] = n/V = 5.0 g NH3 x 1 mol NH3/17.04 g NH3/2.50 L = 0.12 M

            [NH4+] = 20.0 g NH4Cl x 1 mol NH4Cl/53.50 g NH4Cl x 1 mol NH4+/1 mol NH4Cl/2.50 L = 0.150 M

 

                      NH4+(aq) H+(aq)   +   NH3(aq)

            [ ]i       0.150                   0                0.12

            [ ]c         -x                    +x                 +x

            [ ]e      0.150-x                x                0.12+x

 

            Ka = [H+] x [NH3]/[NH4+]

            5.6 x 10-10 = x · (0.12+x)/(0.150-x) ≈ 0.12x/0.150

            [H+] = 7.0 x 10-10 M

            See explanation in previous problem, #14, concerning %ionization calculation.

            pH = -log[H+] = -log(7.0 x 10-10) = 9.15

 

      (b) H+(aq) + NO3-(aq) + NH3(aq) NH4+(aq) + NO3-(aq)

 

      (c) NH4+(aq) + Cl-(aq) + K+(aq) + OH-(aq) NH3(aq) + Cl-(aq) + K+(aq) + H2O(l)