Buffers And Acid-Base Titrations

30) [KOH] = 0.200 M          [HClO₄] = 0.150 M

      V_KOH = 30.0 mL

      KOH(aq)   K⁺(aq) + OH^-(aq)

      HClO₄(aq) H⁺(aq) + ClO₄^-(aq)

      [OH^-] = n/V

      n_OH- = 0.200 mol KOH/L x 1 mol OH^-/1 mol KOH x 30.0 mL x 1 L/10³mL = 6.00 x 10^-3 mol OH^-          

      (a) V_HClO4 = 30.0 mL

            n_H+ = 0.150 mol HClO₄/L x 1 mol H⁺/1 mol HClO₄ x 30.0 mL x 1 L/10³ mL = 4.50 x 10^-3 mol H⁺

                           H⁺(aq)        +        OH^-(aq) H₂O(l)

            n_b    4.50 x 10^-3 mol        6.00 x 10^-3 mol

            n_a                0                  1.50 x 10^-3 mol

            [OH^-] = n_T/V_T = (1.50 x 10^-3 mol OH^-)/(60.0 mL x 1 L/10³ mL) = 0.0250 M

            pOH = -log[OH^-] = -log(0.0250) = 1.600

            pH + pOH = 14.00

            pH = 14.00 - 1.600 = 12.40

      (b) V_HClO4 = 39.5 mL

            n_H+ = 0.150 mol HClO₄/L x 1 mol H⁺/1 mol HClO₄ x 39.5 mL x 1 L/10³ mL = 5.92 x 10^-3 mol H⁺

                           H⁺(aq)        +        OH^-(aq) H₂O(l)

            n_b    5.92 x 10^-3 mol        6.00 x 10^-3 mol

            n_a                0                  8.00 x 10^-5 mol

            [OH^-] = (8.00 x 10^-5 mol OH^-)/(69.5 mL x 1 L/10³ mL) = 1.15 x 10^-3 M

            pOH = -log[OH^-] = -log(1.15 x 10^-3) = 2.939

            pH + pOH = 14.00

            pH = 14.00 - 2.939 = 11.06

      (c) V_HClO4 = 39.9 mL

            n_H+ = 0.150 mol HClO₄/L x 1 mol H⁺/1 mol HClO₄ x 39.9 mL x 1 L/10³ mL = 5.98 x 10^-3 mol H⁺

                           H⁺(aq)        +        OH^-(aq) H₂O(l)

            n_b    5.98 x 10^-3 mol        6.00 x 10^-3 mol

            n_a                0                  2.00 x 10^-5 mol

            [OH^-] = (2.00 x 10^-5 mol OH^-)/(69.9 mL x 1 L/10³ mL) = 2.86 x 10^-4 M

            pOH = -log[OH^-] = -log(2.86 x 10^-4) = 3.544

            pH + pOH = 14.00

            pH = 14.00 - 3.544 = 10.46

      (d) V_HClO4 = 40.0 mL

            n_H+ = 0.150 mol HClO₄/L x 1 mol H⁺/1 mol HClO₄ x 40.0 mL x 1 L/10³ mL = 6.00 x 10^-3 mol H⁺

                           H⁺(aq)        +        OH^-(aq) H₂O(l)

            n_b    6.00 x 10^-3 mol        6.00 x 10^-3 mol

            n_a                0                               0

            K_w = [H⁺] x [OH^-] = 1.00 x 10^-14

            [H⁺] = [OH^-] = 1.00 x 10^-7 M

            pH = -log[H⁺] = -log(1.00 x 10^-7) = 7.000

      (e) V_HClO4 = 40.1 mL

            n_H+ = 0.150 mol HClO₄/L x 1 mol H⁺/1 mol HClO₄ x 40.1 mL x 1 L/10³ mL = 6.02 x 10^-3 mol H⁺

                           H⁺(aq)        +        OH^-(aq) H₂O(l)

            n_b    6.02 x 10^-3 mol        6.00 x 10^-3 mol

            n_a    2.00 x 10^-5 mol                    0                  

            [H⁺] = (2.00 x 10^-5 mol H⁺)/(70.1 mL x 1 L/10³ mL) = 2.85 x 10^-4 M

            pH = -log[H⁺] = -log(2.85 x 10^-4) = 3.588