Ex8 - lab report chm432 PDF

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EXPERIMENT 8: SURFACE CHEMISTRY (Adsorption of acetic acid on activated carbon) OBJECTIVE: To study the adsorption isotherm of acetic acid by activated carbon. INTRODUCTION: Adsorption is the accumulation of a gas or liquid solute (the adsorbate) on a surface of a solid or a liquid (the adsorbent) f...

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EXPERIMENT 8: SURFACE CHEMISTRY (Adsorption of acetic acid on activated carbon) OBJECTIVE: To study the adsorption isotherm of acetic acid by activated carbon. INTRODUCTION: Adsorption is the accumulation of a gas or liquid solute (the adsorbate) on a surface of a solid or a liquid (the adsorbent) forming a molecular or atomic film. It is different from absorption, in which a substance diffuses into a liquid or solid to form a solution. Adsorption has many applications. Its first application was probably the use of bone ashes to remove color from syrups or alcohol. Today, adsorbent materials are widely used at water-treatment plants to remove especially organic impurities and chlorine as purification treatment. Atoms on the surface are not wholly surrounded by other atoms and experience a bond deficiency. Thus it is generally energetically favorable for them to bond to the adsorbate. Adsorption on solid surfaces may be either as chemisorption where a chemical band forms between the surface and the adsorbate - such as the adsorption of chloride ions on AgCl (ionic bond) or of oxygen gas on metals where oxygen to metal bonds are formed (covalent bond) or physical adsorption results from attractions like nonpolar Vander Waals, dipole - dipole etc. The dependence of the extent of adsorption on concentration in bulk is frequently called as the "adsorption isotherm" that is, a relation that describes the amount of adsorbate on the adsorbent as a function of its pressure (if gas) or concentration (if liquid). There are several types of isotherms describing process of adsorption, namely Freundlich isotherm, Langmuir isotherm, BET isotherm, etc. Langmuir isotherm is applied when the adsorption occurs as a single layer (monolayer), where the following equilibrium condition is considered

APPARATUS: 1) 250 cm³ conical flask. 2) 20 cm³ beaker. 3) Burette. 4) 25 cm³ pipette.

5) 100 cm³ measuring cylinder. 6) Filter funnel. 7) Retort stand. 8) Parafilm. CHEMICALS: 1) Activated charcoal (carbon). 2) 0.4M Acetic acid CH3COOH. 3) 0.1M Sodium hydroxide, NaOH. 4) Phenolphthalein indicator. PROCEDURE: 1) 1.5g of activated charcoal was weighted and was placed into six dry 250 mL conical flasks. 2) 100 mL of acetic acid solutions was prepared according table 8.1 by using 100 ml of measuring cylinder to measure the required amount of 0.4 M acetic acid ten was diluted with distilled water up to the mark. Table 8.1: Suggested volumes of 0.4 M acetic acid to be diluted to 100cm³. Sample 1 2 3 4 5 6

Volume 0.4M acetic acid (cm³) 100 75 50 25 10 5

3) For each sample above, 100 mL of the acetic acid solution was added to a charcoal sample. The mixed solutions in the flask was swirled and let them stand for a week. 4) After a week, the solutions were filtered. 5) 2-3 drops of phenolphthalein were added to the filtered solution and was titrated with 0.1 M NaOH (Refer table 8.2). Table 8.2: Volume of filtrate required for analysis of sample. Sample

Volume of filtrate (cm³)

1 2 3 4 5 6

10 10 10 25 25 40

6) The titration process was repeated 3 times for each sample. RESULTS: Sample Mass of activated carbon (g)

1 1.5059

2 1.5115

3 4 1.5094 1.5073

5 1.5097

6 1.507 8

Titration: Volume of NaOH (ml) Trial 1 Trial 2 Trial 3 Average volume of NaOH (ml)

50 35.2 33.8 34.2 34.4

24.8 25.4 26.2 25.5

CALCULATIONS: 1) The initial concentration diluted of acetic acid: a) Sample 1. M1V1 = M2V2 (0.4M) (100mL) = (M2) (100mL) M2 = 0.4 M b) Sample 2. M1V1 = M2V2 (0.4M) (75mL) = (M2) (100mL) M2 = 0.3 M c) Sample 3 M1V1 = M2V2 (0.4M) (50mL) = (M2) (100mL) M2 = 0.2 M d) Sample 4. M1V1 = M2V2

16.0 16.6 16.4 16.3

18.1 18.2 18.4 18.2

6.8 6.5 6.2 6.5

6.0 5.8 5.9

(0.4M) (25mL) = (M2) (100mL) M2 = 0.1 M e) Sample 5. M1V1 = M2V2 (0.4M) (10mL) = (M2) (100mL) M2 = 0.04 M f) Sample 6. M1V1 = M2V2 (0.4M) (5mL) = (M2) (100mL) M2 = 0.02 M 2) The final concentration of diluted acetic acid: a) Sample 1. M1V1 = M2V2 (0.4M) (100mL) = (M2) (250mL) M2 = 0.16 M b) Sample 2. M1V1 = M2V2 (0.4M) (75mL) = (M2) (250mL) M2 = 0.12 M c) Sample 3 M1V1 = M2V2 (0.4M) (50mL) = (M2) (250mL) M2 = 0.08 M d) Sample 4. M1V1 = M2V2 (0.4M) (25mL) = (M2) (250mL)

M2 = 0.04 M e) Sample 5. M1V1 = M2V2 (0.4M) (10mL) = (M2) (250mL) M2 = 0.016 M f) Sample 6. M1V1 = M2V2 (0.4M) (5mL) = (M2) (250mL) M2 = 0.008 M 3) The final (equilibrium) concentration of each acetic solution after titrated with NaOH: a) Sample 1 M1V1 = M2V2 (M2) (100mL) = (0.10M) (34.4mL) M2 = 0.0344 M b) Sample 2 M1V1 = M2V2 (M2) (75mL) = (0.10M) (25.5mL) M2 =0.0340 M c) Sample 3 M1V1 = M2V2 (M2) (50mL) = (0.10M) (16.3mL) M2 = 0.0326 M d) Sample 4

M1V1 = M2V2 (M2) (25mL) = (0.10M) (18.2mL) M2 =0.0728 M e)

Sample 5 M1V1 = M2V2 (M2) (10mL) = (0.10M) (6.5mL) M2 =0.0650 M

f)Sample 6 M1V1 = M2V2 (M2) (40mL) = (0.10M) (5.9mL) M2 =0.1180 M 4) The number of moles of acetic acid adsorbed by 1 g of charcoal. 1. Sample 1: a) Initial mol = n = MV/ 1000 = 0.4 M x 100 mL / 1000 = 0.04 mol b) Final concentration = 0.16 M, Final mol = n = MV/ 1000 = 0.16 M x 100 mL / 1000 = 0.016 mol c) Number of mole of CH3COOH adsorb = Initial mol – final mol = 0.04 mol – 0.016 mol =0.024 mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon

N = number of mole of CH3COOH adsorb Mass of activated carbon = 0.024 mol / 1.5059 g = 0.0159 mol/g 2. Sample 2: a) Initial mol = n = MV/ 1000 = 0.3 M x 100 mL / 1000 = 0.03 mol b) Final concentration = 0.12 M, Final mol = n = MV/ 1000 = 0.12 M x 100 Ml / 1000 = 0.012 mol c) Number of mole of CH3COOH adsorb Initial mol – final mol =0.03 mol – 0.012 mol =0.018 mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon N = number of mole of CH3COOH adsorb Mass of activated carbon = 0.018 mol / 1.5115 g = 0.0119 mol/g

3. Sample 3: a) Initial mol = n = MV/ 1000 = 0.2 M x 100 mL / 1000 = 0.02 mol b) Final concentration = 0.08 M, Final mol = n = MV/ 1000

= 0.08 M x 100 Ml / 1000 = 0.008 mol c) Number of mole of CH3COOH adsorb Initial mol – final mol =0.02 mol – 0.008 mol =0.012 mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon N = number of mole of CH3COOH adsorb Mass of activated carbon = 0.012 mol / 1.5094 g = 7.95 x 10¯³ mol/g 4. Sample 4: a) Initial mol = n = MV/ 1000 = 0.1 M x 100 mL / 1000 = 0.01 mol b) Final concentration = 0.04 M, Final mol = n = MV/ 1000 = 0.04 M x 100 mL / 1000 = 0.004 mol c) Number of mole of CH3COOH adsorb = Initial mol – final mol = 0.01 mol – 0.004 mol =0.006 mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon N = number of mole of CH3COOH adsorb Mass of activated carbon = 0.006 mol / 1.5073 g = 3.98 x 10¯³ mol/g

5. Sample 5: a) Initial mol = n = MV/ 1000 = 0.04 M x 100 mL / 1000 = 0.004 mol b) Final concentration = 0.016 M, Final mol = n = MV/ 1000 = 0.016 M x 100 mL / 1000 = 0.0016 mol c) Number of mole of CH3COOH adsorb = Initial mol – final mol = 0.004 mol – 0.0016 mol =0.0024 mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon N = number of mole of CH3COOH adsorb Mass of activated carbon = 0.0024 mol / 1.5097 g = 1.59 x10¯³ mol/g 6. Sample 6: a) Initial mol = n = MV/ 1000 = 0.02 M x 100 mL / 1000 = 0.002 mol b) Final concentration = 0.008 M, Final mol = n = MV/ 1000 = 0.008 M x 100 mL / 1000 = 0.0008 mol c) Number of mole of CH3COOH adsorb = Initial mol – final mol = 0.002 mol – 0.0008 mol

=1.2 x10¯³ mol d) N = Number of mole of CH3COOH adsorb per gram of activated carbon N = number of mole of CH3COOH adsorb Mass of activated carbon = 1.2 x10¯³ mol / 1.5078 g = 7.96 x 10^¯4 mol/g

3. Plot the Langmuir isotherm, c/y versus c. Sample 1

c 0.4

c/y = (0.4 M) / (1.59 x 10¯² mol/g)

2

0.3

=25.16 g/L = (0.3 M) / (1.19 x 10¯² mol/g)

3

0.2

=25.21 g/L = (0.2 M) / (7.95 x 10¯³mol/g)

4

0.1

=25.16 g/L = (0.1 M) / (3.98 x 10¯³mol/g)

0.04

=25.13 g/L = (0.04 M) / (1.59 x 10¯³mol/g)

0.02

=25.13 g/L = (0.02 M) / (7.96 x 10^-4 mol/g)

5 6

=25.13 g/L...