1. Lab Activity 2: Active Acidity, pH, and Buffer IUG, Fall 2012 Dr. Tarek Zaida IUG, Fall 2012 Dr. Tarek Zaida 1

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. 2

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 3

4. pH-indicator paper It contains organic dye whose color is dependent on pH  pH scale is 1 - 14 4

5. The pH-meter Is an instrument equipped with: 1.A glass electrode 2.A reference electrode (calomel: Hg, HgCl 2 )  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. 5

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. 6

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. 7

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

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

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 5. 0.1 N HCl 6. 0.1 N NaOH 10

11. Experiment 2: Preparation of buffers Prepare the acetate buffer of pH 5.0, keeping in mind that: pK a 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 11

12. pH = pKa + log [A - ]/[HA] [A - ] = x [HA] = 0.1 – x 5 = 4.7 + log ( x / 0.1- x) 0.3 = log (x/ 0.1 – x) 1.995 = x/0.1 – x And x = 0.1995 – 1.995x 2,995 x = 0.1995 X = 0.1995/2.995 X = 0.0666 mole/l CH 3 COONa CH 3 COOH= 0.1 – 0.0666 = 0.0334 mol/l CH 3 COOH 12

13. MW of CH 3 COONa = 82 MW of CH 3 COOH = 60 For 0.0666 mol/l of CH 3 COONa: 0.0666 mol/l x 82 = 5.41 g/l are required For 0.0334 mol/l of CH 3 COOH: 0.0334 mol/l x 60 = 2.004 g/l Note: Density of CH3COOH = 1.05 2.004 / 1.05 = 1.9 ml /l are required 13