Buffers And Acid-Base Titrations

16) (a) m_NH3 = 5.0 g NH₃          m_NH4Cl = 20.0 g NH₄Cl    

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

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

            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

            [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

            K_a = [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)