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.