Buffers And Acid-Base Titrations

 

32) [NH3] = 0.030 M          [HCl] = 0.025 M

      VNH3 = 30.0 mL            Kb(NH3) = 1.8 x 10-5

 

      NH3(aq)  +  H2O(l) NH4+(aq) + OH-(aq)

      HCl(aq) H+(aq) + Cl-(aq)

 

      [NH3] = n/V

      nNH3 = 0.030 mol NH3/L x 30.0 mL x 1 L/103 mL = 9.0 x 10-4 mol NH3

 

      (a) VHCl = 0 mL

            

                    NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

            [ ]i      0.030                                0                 0

            [ ]c         -x                                +x               +x           

            [ ]e     0.030-x                             x                 x

 

            Kb = [NH4+] x [OH-]/[NH3]

            1.8 x 10-5 = x • x/(0.030 - x) ≈ x2/0.030

            [OH-] = 7.3 x 10-4 M

            pOH = -log[OH-] = -log(7.3 x 10-4) = 3.14

            pH + pOH = 14.00

            pH = 14.00 - 3.14 = 10.86

     

      (b) VHCl = 10.0 mL

 

            nH+ = 0.025 mol HCl/L x 1 mol H+/1 mol HCl x 10.0 mL x 1 L/103 mL = 2.5 x 10-4 mol H+

            

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

            nb    2.5 x 10-4 mol     9.0 x 10-4 mol              0

            na           0                  6.5 x 10-4 mol        2.5 x 10-4 mol

 

            [NH3] = (6.5 x 10-4 mol NH3)/(40.0 mL x 1 L/103 mL) = 0.016 M

            [NH4+] = (2.5 x 10-4 mol NH4+)/(40.0 mL x 1 L/103 mL) = 0.0062 M

 

                    NH3(aq) + H2O(l) NH4+(aq)   +   OH-(aq)

            [ ]i      0.016                              0.0062              0

            [ ]c         -x                                    +x               +x           

            [ ]e     0.016-x                           0.0062+x          x

 

            Kb = [NH4+] x [OH-]/[NH3]

            1.8 x 10-5 = (0.0062 + x) • x/(0.016 - x) ≈ 0.0062x/0.016

            [OH-] = 4.6 x 10-5 M

            pOH = -log[OH-] = -log(4.6 x 10-5) = 4.34

            pH + pOH = 14.00

            pH = 14.00 - 4.34 = 9.66

 

      (c) VHCl = 20.0 mL

 

            nH+ = 0.025 mol HCl/L x 1 mol H+/1 mol HCl x 20.0 mL x 1 L/103 mL = 5.0 x 10-4 mol H+

            

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

            nb    5.0 x 10-4 mol     9.0 x 10-4 mol              0

            na           0                  4.0 x 10-4 mol        5.0 x 10-4 mol

 

            [NH3] = (4.0 x 10-4 mol NH3)/(50.0 mL x 1 L/103 mL) = 0.0080 M

            [NH4+] = (5.0 x 10-4 mol NH4+)/(50.0 mL x 1 L/103 mL) = 0.010 M

 

                    NH3(aq) + H2O(l) NH4+(aq)   +   OH-(aq)

            [ ]i      0.0080                             0.010               0

            [ ]c         -x                                    +x               +x           

            [ ]e     0.0080-x                          0.010+x           x

 

            Kb = [NH4+] x [OH-]/[NH3]

            1.8 x 10-5 = (0.010 + x) • x/(0.0080 - x) ≈ 0.010x/0.0080

            [OH-] = 1.4 x 10-5 M

            pOH = -log[OH-] = -log(1.4 x 10-5) = 4.85

            pH + pOH = 14.00

            pH = 14.00 - 4.85 = 9.15

          

      (d) VHCl = 35.0 mL

 

            nH+ = 0.025 mol HCl/L x 1 mol H+/1 mol HCl x 35.0 mL x 1 L/103 mL = 8.8 x 10-4 mol H+

            

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

            nb     8.8 x 10-4 mol     9.0 x 10-4 mol             0

            na           0                   2.0 x 10-5 mol        8.8 x 10-4 mol

 

            [NH3] = (2.0 x 10-5 mol NH3)/(65.0 mL x 1 L/103 mL) = 3.1 x 10-4 M

            [NH4+] = (8.8 x 10-4 mol NH4+)/(65.0 mL x 1 L/103 mL) = 0.014 M

 

                    NH3(aq) + H2O(l) NH4+(aq)   +   OH-(aq)

            [ ]i    3.1 x 10-4                          0.014               0

            [ ]c         -x                                    +x               +x           

            [ ]e     3.1 x 10-4-x                     0.014+x           x

 

            Kb = [NH4+] x [OH-]/[NH3]

            1.8 x 10-5 = (0.014 + x) • x/(3.1 x 10-4 - x) ≈ 0.014x/(3.1 x 10-4)

            [OH-] = 4.0 x 10-7 M

            pOH = -log[OH-] = -log(4.0 x 10-7) = 6.40

            pH + pOH = 14.00

            pH = 14.00 - 6.40 = 7.60

                  

      (e) VHCl = 36.0 mL

 

            nH+ = 0.025 mol HCl/L x 1 mol H+/1 mol HCl x 36.0 mL x 1 L/103 mL = 9.0 x 10-4 mol H+

            

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

            nb     9.0 x 10-4 mol     9.0 x 10-4 mol             0

            na           0                          0                     9.0 x 10-4 mol

 

            [NH4+] = (9.0 x 10-4 mol NH4+)/(66.0 mL x 1 L/103 mL) = 0.014 M

 

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

            [ ]i       0.014                              0                  0

            [ ]c         -x                               +x                +x           

            [ ]e      0.014-x                           x                  x

 

            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

            Ka = [H+] x [NH3]/[NH4+]

            5.6 x 10-10 = x • x/(0.014-x) ≈ x2/0.014

            [H+] = 2.8 x 10-6 M

            pH = -log[H+] = -log(2.8 x 10-6) = 5.55

            

      (f) VHCl = 37.0 mL

 

            nH+ = 0.025 mol HCl/L x 1 mol H+/1 mol HCl x 37.0 mL x 1 L/103 mL = 9.2 x 10-4 mol H+

            

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

            nb     9.2 x 10-4 mol     9.0 x 10-4 mol             0

            na     2.0 x 10-5 mol           0                     9.0 x 10-4 mol

 

            [H+] = 2.0 x 10-5 mol H+/(67.0 mL x 1 L/103 mL) = 3.0 x 10-4 M

            pH = -log[H+] = -log(3.0 x 10-4) = 3.52