1. Lab Activity 1: Active Acidity, pH, and Buffer
IUG, Fall 2017
Dr. Tarek Zaida

2. Active Acidity
Refers to H+ present in a solution due to dissociation of acid.
Q: How can active acidity be expressed?
An: by pH of a solution.

3. pH presents the negative logarithm of the hydrogen ion conc. [H+]
pH = -log [H+]
If [H+] = a x 10-b moles/l
Then pH = b – log (a)
pH of a solution can be measured in 2 different ways:
pH-meter
pH-indicator paper

4. pH-indicator paper
It contains organic dye whose color is dependent on pH
pH scale is

5. The pH-meter Is an instrument equipped with: A glass electrode
A reference electrode (calomel: Hg, HgCl2)
How does it work?
The instrument measures the potential difference between the glass and the calomel (mercury & mercury chloride) electrodes.
The potential difference is related to [H+] of the solution being tested.

6. Standardization of pH-meter
All pH-meters should be standardized with buffers of known pH-values before use:
pH4, pH7, pH10
What is a buffer?
Solutions made of a mixture of a week acid & it’s conjugate base.

7. Function of buffers
They are of a vital importance by their ability to maintain the optimal pH in enzyme-catalysed reactions in vitro or in vivo.
How do buffers function in a solution?
They can release H+ in solution if the pH gets basic or
They can bind H+ if the pH gets acidic.

8. HA H+ + A- Ka , a dissociation constant: Ka = [H+] [A-] [HA] pH = pKa + log [A-] Henderson-Hasselbalch equation If any 2 parts of the equation are known, the third can be calculated.

9. If [A-] equal [HA]
Then pH = pKa + log [A-]
[HA]
Then pH = pKa

10. Experiment 1: Measurement of pH
Determine the pH of the following solutions using a pH-meter:
1. Orange or lemon juice
2. Vinegar
3. Distilled water
4. Distilled water boiled
N HCl
N NaOH

11. Experiment 2: Preparation of buffers
Prepare the acetate buffer of pH 5.0, keeping in mind that:
pKa of acetic acid is: 4.7 at pH 5.0
The conjugate base of acetic acid is: CH3COO-
Calculations: for using Henderson-Hasselbalch equation:
pH is given
pKa is also given

12. pH = pKa + log [A-]/[HA] [A-] = x
[HA] = 0.1 – x
5 = log ( x / 0.1- x)
0.3 = log (x/ 0.1 – x)
= x/0.1 – x
And x = – 1.995x
2,995 x =
X = /2.995
X = mole/l CH3COONa
CH3COOH= 0.1 – = mol/l CH3COOH

13. MW of CH3COONa = 82 MW of CH3COOH = 60 For mol/l of CH3COONa: mol/l x 82 = 5.41 g/l are required For mol/l of CH3COOH: mol/l x 60 = g/l Note: Density of CH3COOH = / 1.05 = 1.9 ml /l are required