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

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.

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

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

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.

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.

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.

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.

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

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

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

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 CH3COONa CH3COOH= 0.1 – 0.0666 = 0.0334 mol/l CH3COOH

MW of CH3COONa = 82 MW of CH3COOH = 60 For 0.0666 mol/l of CH3COONa: 0.0666 mol/l x 82 = 5.41 g/l are required For 0.0334 mol/l of CH3COOH: 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