Buffers And Acid-Base Titrations

 

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

      VKOH = 30.0 mL

 

      KOH(aq)   K+(aq) + OH-(aq)

      HClO4(aq) H+(aq) + ClO4-(aq)

 

      [OH-] = n/V

      nOH- = 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) VHClO4 = 30.0 mL

            nH+ = 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)

            nb    4.50 x 10-3 mol        6.00 x 10-3 mol

            na                0                  1.50 x 10-3 mol

 

            [OH-] = nT/VT = (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) VHClO4 = 39.5 mL

            nH+ = 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)

            nb    5.92 x 10-3 mol        6.00 x 10-3 mol

            na                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) VHClO4 = 39.9 mL

            nH+ = 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)

            nb    5.98 x 10-3 mol        6.00 x 10-3 mol

            na                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) VHClO4 = 40.0 mL

            nH+ = 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)

            nb    6.00 x 10-3 mol        6.00 x 10-3 mol

            na                0                               0

 

            Kw = [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) VHClO4 = 40.1 mL

            nH+ = 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)

            nb    6.02 x 10-3 mol        6.00 x 10-3 mol

            na    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