E785: Acid/Base – Effectiveness of a Buffer

Introduction

Strong base is added to three different buffers and water and the effect on pH is observed.

Buffer 1


                             pH = pKa - log([HOAc]/[OAc-])

                                  = 4.74 - log(1/1)  

                                  =  4.74

            After addition of 2 mL 1 M NaOH:                              

                   HOAc         +           OH-             ——>             H2O           +          OAc-

                 0.10 mol             0.0020 mol                                                       0.10 mol

              -0.0020 mol         -0.0020 mol                                                   +0.0020 mol

                0.098 mol                 0 mol                                                           0.102 mol

                                    new pH   = pKa - log{(0.098 mol/0.102 L)/(0.102 mol/0.102 L)

                                                   = pKa - log (0.098/0.102) = 4.74 - log(0.961)

                                                   = 4.76

                                    DpH        = +0.02

            After addition of 10 mL 1 M NaOH:                             

                   HOAc         +           OH-             ——>             H2O           +          OAc-

                 0.10 mol                0.01 mol                                                         0.10 mol

                -0.01 mol              -0.01 mol                                                        +0.01 mol

                 0.09 mol                  0 mol                                                            0.11 mol

                                    new pH   = pKa - log{(0.09 mol/0.11 L)/(0.11 mol/0.11 L)

                                                   = pKa - log (0.09/0.11) = 4.74 - log(0.818)

                                                   = 4.83

                                    DpH        = +0.09

Buffer 2


                             pH          = pKa - log([HOAc]/[OAc-])

                                                   = 4.74 - log(0.001/1)

                                                   = 7.74

            After addition of 2 mL 1 M NaOH:

                                   

                   HOAc         +           OH-             ——>             H2O           +          OAc-

               0.0001 mol           0.0020 mol                                                       0.10 mol

              -0.0001 mol          -0.0001 mol                                                   +0.0001 mol

                    0 mol                0.0019 mol                                                     0.1001 mol

          Solution is no longer a buffer; pH comes from excess OH- after HOAc is consumed.

                                    pOH       = -log (0.0019 mol/0.102) = 1.73

                                    pH          = 12.27

                                    DpH        = +4.53

Buffer 3


                             pH = pKa - log([HOAc]/[OAc-]) = 4.74 - log(0.1/0.1)  =  4.74

            After addition of 2 mL 1 M NaOH:

                                  

                   HOAc         +           OH-             ——>             H2O           +          OAc-

                0.010 mol             0.002 mol                                                       0.010 mol

               -0.002 mol            -0.002 mol                                                     +0.002 mol

                0.008 mol                 0 mol                                                           0.012 mol

                                    new pH   = pKa - log{(0.008 mol/0.102 L)/(0.012 mol/0.102 L)

                                                   = pKa - log (0.008/0.012) = 4.74 - log(0.667)

                                                   = 4.92

                                    DpH        = +0.18

            After addition of 10 mL 1 M NaOH:

                              

                   HOAc         +           OH-             ——>             H2O           +          OAc-

                0.010 mol             0.010 mol                                                       0.010 mol

               -0.010 mol            -0.010 mol                                                     +0.010 mol

                    0 mol                     0 mol                                                           0.020 mol

                        Solution is no longer a buffer; pH comes from hydrolysis of weak base:

                                                                            Kb

                     OAc            +          H2O               ——>          OH-           +         HOAc

                 0.182 M                                                              0 M                        0 M

                   -x                                                                       +x                         +x           

                0.182-x M                                                             x M                        x M

                                    Kb = 5.56x10-10 = (x)(x)/(0.182)

                                       x = [OH-] = 1.01x10-5 M      pOH = 5.00

                                    pH          = 9.00

                                    DpH        = +4.26

Water

                             pH          = 7.00

                                    After addition of 2 mL 1 M NaOH:

                                    pOH       = -log (0.0020/0.102) = 1.71

                                    pH          = 12.29

                                    DpH        = +5.29

 To Conduct Demonstration:

1.    Place 150 mL beakers containing 100 mL of each of three buffers and distilled water on the overhead projector.  Add a few drops of universal indicator so that the color is sufficiently intense.

                                           Buffer                             pH                             Color       

                                               1                               4.74                          orange

                                               2                                7.9                             green

                                               3                               4.69                          orange

                                           Water                            5.5                            orange

2.  Add 2 mL of 1 M NaOH to each beaker.  Results should be:

  1.  Water goes from orange to blue.
  2.  Buffers 1 and 3 remain orange; equimolar concentrations of acid and conjugate base withstand small DpH.
  3. Buffer 2 goes from green to blue; 1000:1 HOAc:OAc- ratio is ineffective buffer, pH goes from 7.9 to 12.8.

3.  Add 8 mL of 1 M NaOH to buffer 1 and buffer 3.

  1.  Buffer 1 remains orange; large (1 M) equimolar concentrations of acid and conjugate base give large buffer  capacity.
  2. Buffer 3 goes from orange to blue; smaller (0.1 M) concentrations of acid and conjugate base lead to lower buffer capacity than A.

 Safety and Disposal

Goggles should be worn. 

All solutions may be sink-disposed.

Reference:

S.S. Zumdahl, Chemistry, pp. 644-650, 1989.