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Lab Activity 1: Active Acidity, pH, and Buffer
IUG, Dr. Tarek Zaida
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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.
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pH value 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
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pH-indicator paper It contains organic dye whose color is dependent on pH pH scale is
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The pH-meter
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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.
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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.
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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.
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For the dissociation of a weak acid:
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.
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For the dissociation of a weak acid:
If [A-] equal [HA] Then pH = pKa + log [A-] [HA] Then pH = pKa
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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
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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
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Experiment 2: Preparation of buffers
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
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Experiment 2: Preparation of buffers
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
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