Buffers And Acid-Base Titrations

14) (a) [NaHCO₃] = 0.100 M          [Na₂CO₃] = 0.125 M          Ka(HCO₃^-) = 5.6 x 10^-11

            NaHCO₃(aq) Na⁺(aq) + HCO₃^-(aq)

            Na₂CO₃(aq) 2Na⁺(aq) + CO₃^2-(aq)

            [HCO₃^-] = 0.100 M NaHCO₃ x 1 mol HCO₃^-/1 mol NaHCO₃ = 0.100 M

            [CO₃^2-] = 0.125 M NaHCO₃ x 1 mol CO₃^2-/1 mol NaHCO₃ = 0.125 M

                     HCO₃^-(aq) H⁺(aq)   +   CO₃^2-(aq)

            [ ]_i        0.100                    0               0.125

            [ ]_c          -x                      +x                +x

            [ ]_e       0.100 - x                x              0.125 + x

            K_a = [H⁺] x [CO₃^2-]/[HCO₃^-]

            5.6 x 10^-11 = x · (0.125 + x)/(0.100 - x) ≈ 0.125x/0.100

            [H⁺] = 4.5 x 10^-11 M

            %ion = [H⁺]/[HCO₃^-] x 100% = (4.5 x 10^-11 M)/(0.100 M) x 100% = 4.5 x 10^-11%

            Therefore, the assumption that 0.125 + x ≈ 0.125 and 0.100 - x ≈ 0.100 is valid.

            The presence of the conjugate base(acid) of a weak acid(base), usually ensures the 5% rule applies in buffer solutions unless the buffer capacity has been exceeded.

             pH = -log[H⁺] = - log(4.5 x 10^-11) = 10.35

      (b) [NaHCO₃] = 0.20 M          [Na₂CO₃] = 0.15 M          Ka(HCO₃^-) = 5.6 x 10^-11

           V₁ = 55 mL                          V₂ = 65 mL           

            NaHCO₃(aq) Na⁺(aq) + HCO₃^-(aq)

            Na₂CO₃(aq) 2Na⁺(aq) + CO₃^2-(aq)

            [HCO₃^-] = n/V

            n₁ = 0.20 mol HCO₃^-/L x 55 mL x 1 L/10³ mL = 0.011 mol HCO₃^-

            n₂ = 0.15 mol CO₃^2-/L x 65 mL x 1 L/10³ mL = 0.0098 mol CO₃^2-

            [HCO₃^-] = n/V_T = 0.011 mol HCO₃^-/(120 mL x 1 L/10³ mL) = 0.092 M

            [CO₃^2-] = 0.0098 mol CO₃^2-/(120 mL x 1 L/10³ mL) = 0.082 M

                     HCO₃^-(aq) H⁺(aq) + CO₃^2-(aq)

            [ ]_i        0.092                    0           0.082

            [ ]_c          -x                      +x             +x

            [ ]_e       0.092 - x                x         0.082 + x

            K_a = [H⁺] x [CO₃^2-]/[HCO₃^-]

            5.6 x 10^-11 = x · (0.082 + x)/(0.092 - x) ≈ 0.082x/0.092

            [H⁺] = 6.3 x 10^-11

            pH = -log[H⁺] = - log(6.3 x 10^-11) = 10.20