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:
- Water goes from orange to blue.
- Buffers 1 and 3 remain orange; equimolar concentrations of acid and conjugate base withstand small DpH.
- 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.
- Buffer 1 remains orange; large (1 M) equimolar concentrations of acid and conjugate base give large buffer capacity.
- 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.