buffTi32

Buffers And Acid-Base Titrations

 

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

      V_NH3 = 30.0 mL            K_b(NH₃) = 1.8 x 10^-5

 

      NH₃(aq)  +  H₂O(l) NH₄⁺(aq) + OH^–(aq)

      HCl(aq) H⁺(aq) + Cl^–(aq)

 

      [NH₃] = n/V

      n_NH3 = 0.030 mol NH₃/L x 30.0 mL x 1 L/10³ mL = 9.0 x 10^-4 mol NH₃

 

      (a) V_HCl = 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

 

            K_b = [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) V_HCl = 10.0 mL

 

            n_H+ = 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)

            n_b    2.5 x 10^-4 mol     9.0 x 10^-4 mol              0

            n_a           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

 

            K_b = [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) V_HCl = 20.0 mL

 

            n_H+ = 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)

            n_b    5.0 x 10^-4 mol     9.0 x 10^-4 mol              0

            n_a           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

 

            K_b = [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) V_HCl = 35.0 mL

 

            n_H+ = 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)

            n_b     8.8 x 10^-4 mol     9.0 x 10^-4 mol             0

            n_a           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

 

            K_b = [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) V_HCl = 36.0 mL

 

            n_H+ = 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)

            n_b     9.0 x 10^-4 mol     9.0 x 10^-4 mol             0

            n_a           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

 

            K_w = K_a x K_b = 1.00 x 10^-14

            K_a = (1.00 x 10^-14)/(1.8 x 10^-5) = 5.6 x 10^-10

            K_a = [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) V_HCl = 37.0 mL

 

            n_H+ = 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)

            n_b     9.2 x 10^-4 mol     9.0 x 10^-4 mol             0

            n_a     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