1. EXPERIMENT (5) Preparation and Properties of Buffer Solution

2. Theory:
Buffer solution:
A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its salt or weak base and its salt.
Buffer solutions are resistant to PH changes caused by the addition of an acid or base.

3. Examples of buffer solutions: Weak acid & its salt:
Theory:
Examples of buffer solutions:
Weak acid & its salt:
Acetic acid & sodium acetate (CH3COOH) & (CH3COONa)
CH3COOH CH3COO- + H+
CH3COONa CH3COO- + Na+
HCl + (HA + A-)
NaOH + (HA + A-)
2HA + Cl-
2A- + Na+ + H2O

4. Examples of buffer solutions: Weak base & its salt:
Theory:
Examples of buffer solutions:
Weak base & its salt:
Ammonia & ammonium chloride (NH3) & (NH4Cl)
NH3 + H2O NH+4 + OH-
NH4Cl NH+4 + Cl-
HCl + (BOH + B+)
NaOH + (BOH + B+)
H2O + 2B+ + Cl-
Na+ + 2 BOH

5. Procedure: Prepare the following four solution:
Calculate the theoretical PH value for solution.
Measure the PH value for each solution by using PH meter and compare the results with theoretical values.
Solution 1 2 3 4
0.1 M
Sodium acetate 25
Acetic acid 75
125
175
Total volume 50
100
150
200

6. Procedure:
To 25 ml of one of the above buffer solutions, add 5 drops of 0.1 M NaOH then measure the PH of the resulting solution.
To 25 ml of one of the above buffer solutions, add 5 drops of 0.1 M HCl then measure the PH of the resulting solution.
Measure the PH of the distilled water.
To 25 ml sample of water, add 5 drops of 0.1 M NaOH, and to another 25 ml add 5 drops of 0.1 M HCl then – after mixing-measure the PH of the two solutions.

7. Results: Tabulate your results as following: Solution PH Practical
Theoretical
Practical
Effect of HCl
Effect of NaOH 1 2 3 4
Distilled water

8. PH= Pka + log [ salt] [acid]
Calculation:
PH= Pka + log [ salt] [acid]
Where:
ka(acetic acid) = 1.82X10-5
Pka = -log ka = -log (1.82X10-5) = 4.74
[salt] = ( M salt x V salt ) V total [acid] = ( M acid x V acid ) V total

9. PH= Pka + log [ salt] [acid]
Calculation:
Theoretical PH:
PH= Pka + log [ salt] [acid]
(1) → PH= log (0.1 x 25)/50 (0.1 x 25)/50 = = 4.74
(2) → PH= log (0.1 x 25)/100 (0.1 x 75)/100 = = 4.26
(3) → PH= log (0.1 x 25)/150 (0.1 x 125)/150 = = 4.04
(4) → PH= log (0.1 x 25)/200 (0.1 x 175)/200 = = 3.89