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Lab Activity 2:Active Acidity, pH, and Buffer

IUG, Fall 2012Dr. 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 presents the negative logarithm of the hydrogen ion conc. [H+]

pH = -log [H+] If [H+] = a x 10-b moles/lThen pH = b – log (a)pH of a solution can be measured in 2 different

ways:pH-meterpH-indicator paper

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pH-indicator paper

• It contains organic dye whose color is dependent on pH

pH scale is 1 - 14

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The pH-meter

• Is an instrument equipped with:1. A glass electrode2. 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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HA H+ + A-

Ka , a dissociation constant:

Ka = [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.

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• 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• 5. 0.1 N HCl• 6. 0.1 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 givenpKa is also given

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• 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 – xAnd x = 0.1995 – 1.995x2,995 x = 0.1995X = 0.1995/2.995X = 0.0666 mole/l CH3COONa

CH3COOH= 0.1 – 0.0666 = 0.0334 mol/l CH3COOH

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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 requiredFor 0.0334 mol/l of CH3COOH:

0.0334 mol/l x 60 = 2.004 g/lNote: Density of CH3COOH = 1.052.004 / 1.05 = 1.9 ml /l are required