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) + CO32-(aq)

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

            [CO32-] = 0.125 M NaHCO3 x 1 mol CO32-/1 mol NaHCO3 = 0.125 M

 

                     HCO3-(aq) H+(aq)   +   CO32-(aq)

            [ ]i        0.100                    0               0.125

            [ ]c          -x                      +x                +x

            [ ]e       0.100 - x                x              0.125 + x

 

            Ka = [H+] x [CO32-]/[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) + CO32-(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 CO32-/L x 65 mL x 1 L/103 mL = 0.0098 mol CO32-

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

            [CO32-] = 0.0098 mol CO32-/(120 mL x 1 L/103 mL) = 0.082 M

 

                     HCO3-(aq) H+(aq) + CO32-(aq)

            [ ]i        0.092                    0           0.082

            [ ]c          -x                      +x             +x

            [ ]e       0.092 - x                x         0.082 + x

 

            Ka = [H+] x [CO32-]/[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