Exp 4.docx

INTRODUCTION : In liquid-liquid extraction, components in the fed material, consisting of liquid phases are separated when third liquid also known as solvent is added to the process. By adding this new component which is insoluble in the feed, a new phase is formed. The component which is more important during the extraction or which is the desired component to be extracted during the process is transferred to extract. The remaining liquid from the feed which is separable goes to raffinate (fig.1). The selection of solvent is one of the important issues. During the process, two liquid phases are formed and both the components are differently distributed in solvent and extract phases. The difference in phase compositions makes the mass transfer easy. This distribution of the components in aqueous and organic phase is governed by the distribution coefficient denoted by K. The distribution coefficient gives a quantitative measure of how an organic compound will distribute between aqueous and organic phases. K is the ratio of the solubility of solute dissolved in the organic phase to the solubility of solute dissolved in aqueous phase. This means higher the value of K, higher the solute will reside in the organic phase. If the separable components’ boiling points vary, it makes the extraction beneficial. The difference in density can make extraction easier, meaning; higher the difference in density between the phases the extraction process is quicker.

OBJECTIVES : 1. To perform a single-stage extraction to separate acetic acid from its solution in chloroform using water as an extraction solvent. 2. To determine the acid contents in the raffinate and extract phases by titration and its result are compared to the theoretical result.

PROCEDURES : 1.

A binary acetic acid-chloroform solution was prepared by adding 4 mL acetic acid to 5 mL chloroform in separating funnel. (This might generate a lot of vapor. To avoid sudden pressure burst, do not stopper the funnel.)

2. With the funnel mouth still open, the mixture was gently swirl to mix it well into a homogenous binary solution. this was the feed solution.

3. Now the acid was extracted from binary solution using equal mass of an extraction solvent, water. 10 mL distilled water was measured and poured it into the funnel. (This mixture might generate a lot of vapor.) 4. Hand glove was put on. With the stopper on, the funnel was gently shook and then its stopcock was opened to release any pressure build-up. Shaking and venting was continued for about 5 minutes. 5. The funnel was put on the o-ring support and its contents were let settle into phases. 5-10 minutes was waited to allow the heavy and light phases reached their equilibrium. 6. Two clean empty conical flasks were weighed and then each phases was transferred into each flask. The beaker was reweighed. 7. About two drops of phenolphthalein was put into each flask and the solution was titrated using 2.0 M NaOH (aq). The endpoint was when the solution turns color.

RESULT AND DATA : 1) Mass of empty conical flask 1

= 130.2321 g

2) Mass of empty conical flask 1 + light phase

= 143.1171 g

3) Mass of empty conical flask 2

= 128.4532 g

4) Mass of empty conical flask 2 + heavy phase

= 133.2133 g

5) Volume of acetic acid

= 4 mL

6) Volume of chloroform

= 5 mL

7) Volume of distilled water

= 10 mL

8) Density of acetic acid

= 1.049 g/mL

9) Density of chloroform

= 1.843 g/mL

10) Density of water

= 1.000 g/mL

Table1 : Results for titration of phases with 2M NaOH Mass (g)

Volume of 2.0 M NaOH to titrate (mL)

Acetic acid

Chloroform

Water

Heavy

Light phase

Heavy phase

Light phase

12.8850

3.0

27.0

phase 4.196

7.415

10.000

4.7601

CALCULATION : 1. Mass of feed and solvent solution: Feed solution: (4ml of acetic acid x 1.049 g/ml) + (5ml of chloroform x 1.483 g/ml) = 4.196g + 7.415g = 11.611g Solvent solution: 10ml of water x 1.000 g/ml = 10g 2. Mass fraction, XF in feed solution: Mass fraction of acetic acid, XF: = =

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑚𝑖𝑥𝑡𝑢𝑟𝑒 4.196 𝑔 11.611 𝑔

= 0.3614 3. The value of xM :

Total material balance: F+S=M 11.611 + 10 = M M = 21.611

Material balance xM: XF (F) + YS (S) = M (XM) 0.3614 (11.611g) + 0 (10) = 21.611g (XM) XM = 0.1942

4. Mass fractions of acid in raffinate and extract phase: From the equation, CH3 COOH + NaOH → CH3 COONa + H2 O 1 mol of CH3COOH = 1 mol of NaOH

a) In raffinate (heavy phase)

i)

Mol of CH3COOH = mol of NaOH =

𝑀𝑉 1000

=

2(3) 1000

= 0.006 mol ii)

Mass of CH3COOH = mol of CH3COOH x RMM CH3COOH = 0.006 mol X 60 g/mol = 0.36g

iii)

Mass fraction of CH3COOH, x* =

𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝐻3𝐶𝑂𝑂𝐻 𝑚𝑎𝑠𝑠 𝑜𝑓 ℎ𝑒𝑎𝑣𝑦 𝑝ℎ𝑎𝑠𝑒 𝑙𝑖𝑞𝑢𝑖𝑑

=

0.36 𝑔 4.7601 𝑔

= 0.0756

b) In extract (light phase)

i)

Mol of CH3COOH = mol of NaOH 𝑀𝑉

= 1000 =

2(27) 1000

= 0.054 mol

ii)

Mass of CH3COOH = mol of CH3COOH x RMM CH3COOH = 0.054 mol X 60 g/mol = 3.24 g

iii)

Mass fraction of CH3COOH, y* =

𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝐻3𝐶𝑂𝑂𝐻 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑝ℎ𝑎𝑠𝑒 𝑙𝑖𝑞𝑢𝑖𝑑

=

3.24 𝑔 12.8850 𝑔

= 0.2515

DISCUSSION : Liquid-Liquid extraction is a method by which a compound is pulled from solvent A to solvent B where solvents A and B are not miscible. Extraction methods differ depending upon the density of the solvent being used. In this experiment, from values of volume of 2.0 M NaOH to titrate (mL) , the mol of CH3COOH in light phase and heavy phase can be found. Then, mass of CH3COOH and mass fraction of CH3COOH of y* and x* can be calculated. The mass fraction of CH3COOH in raffinate, x* was 0.0756 and the mass fraction of CH3COOH in extract, y* was 0.2515. From the graph,using XF = 0.3614, XM = 0.1942, then theoretical mass fraction in raffinate and extract can be obtained. The theoretical mass fraction of CH3COOH in raffinate, x* was 0.08 and the theoretical mass fraction of CH3COOH in extract, y* was 0.26. From comparison between the theoretical and experimental of mass fraction of raffinate, x* and mass fraction of extract, y*, the value of experimental was almost same with theoretical value. Therefore, this experiment can be categorized as successful experiment. Although the experimental mass fraction of raffinate and extract almost same with theoretical mass fraction of raffinate and extract, there was still a small difference because of some errors that might happen during conducting experiment. There might be errors such as when conducting the venting process, the gas pressure might not be removed properly or when conducting titration, the eyes might not parallel to meniscus of the burette. Other than that, there might be some error during weighing the solute, solvent or solution. All of this errors might disturb the results of the experiment. Although distillation and extraction are two of the most commonly used physical separation methods having an equal importance in the industry to obtain pure chemicals for many applications, there exists a difference between distillation and extraction based on their procedures. The key difference between distillation and extraction is that distillation follows heating of a liquid mixture and collecting the vapor of the liquid at their boiling point and condensing the vapor to get the pure substance whereas, in extraction, a suitable solvent is used for the separation process. Distillation: This separation method is used in the fractional distillation of crude oil production, chemical and petroleum industry. For example, to separate benzene from toluene, ethanol or methanol from water and acetic acid from acetone. Extraction: It is used to isolate organic compounds such as phenol, aniline and nitrated aromatic compounds from water. It is also useful to extract essential oils, pharmaceuticals, flavors, fragrances and food products.

.

CONCLUSION : In conclusion, a single-stage extraction to separate acetic acid from its solution in chloroform using water as an extraction solvent was performed. Then, the acid contents in the raffinate and extract phases was determined by titration and its result are compared to the theoretical result. The mass fraction of CH3COOH in raffinate, x* was 0.0756 and the mass fraction of CH3COOH in extract, y* was 0.2515. The theoretical mass fraction of CH3COOH in raffinate, x* was 0.08 and the theoretical mass fraction of CH3COOH in extract, y* was 0.26. From comparison between the theoretical and experimental of mass fraction of raffinate, x* and mass fraction of extract, y*, the value of experimental was almost same with theoretical value. REFERENCES : 

http://facstaff.cbu.edu/rprice/lectures/extract.html



https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techni ques/Liquid-Liquid_Extraction



http://www.thermopedia.com/content/752/

Appendix 2: Ternary Diagram of Acetic Acid-Chloroform-Water at 1atm