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acid base extraction lab report for organic chemistry 1...

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Acid Base Extraction

Writer: Cas Stone Reviewer: Wafa Tamimi Editor: Amal Tamimi

Introduction Liquid- Liquid eExtraction is a technique used to separate organic compounds from a mixture of solutions. This usually done by using a separatory funnel (3). An example of this could be when making tea or coffee. When solid coffee beans or tea leaves are inserted into hot water (solvent), the soluble flavors or caffeine is then extracted (3). It is commonly used to extract natural compounds, such as essential oils. Fruit leaves are mostly composed of cellulose and water, but also contain oils that can be extracted and used to scent various products.1 ExtractionIt relies on 3 different concepts: molecular polarity and density, and “like dissolves like”. Polar solvents tend to dissolve polar compounds and nonpolar solvents tend to dissolve nonpolar compounds. Molecules with intermediate polarity can be dissolved in either polar or nonpolar solvents. Utilizing the “like dissolves like” method will help in determining which solvent to use for accurate separation. When liquids that do not mix are combined, a top and bottom layer are formed. These are called the “aqueous” layer and the “organic” layer and will allow the removal of one layer at a time (1). The bottom layer is usually occupied by the higher density solvent. For these layers to occur, the two compounds in the mixture must differ greatly in polarity. This separation allows for easy retrieval of the products forof the extraction. It is very important to note that the extraction process must be completed more than once to completely separate the compounds. It is recommended to repeat the process at least 3 time to allow for correct separation. An acid-base extraction is a type of liquid-liquid extraction. It usually involves different solubility levels in water and the organic solvent being used (2). The main difference between the two extractions is that acid-base extractions is use acid-base properties to separate compounds from each other (2). The most important part of an acid-base extraction is identifying sites of acidity and basicity in the target molecule. In this acid base extraction, HCl is added to the basic compound to protonate it and make it dissolve in water. When Bronsted-Lowry acids and bases react, the acid donates a proton to the base, giving the acid a negative charge and the base a positive charge. By making the base polar, it is able to be separated from the nonpolar organic compounds. The protonated base, or conjugate acid, can be removed from the aqueous solution by adding a base to the solution (1). The protonated base will donate a hydrogen to the base that was added and it will become nonpolar again, causing it to fall out of the aqueous solution. Because the nonpolar organic compound is not completely insoluble in water, it is usually necessary to “dry” the organic solution and remove any residual water. This is done by adding an ionic solid such as sodium sulfate. Performing this experiment will apply techniques such as determining the location of organic and aqueous layers based on density, determining whether a substance will act as a Bronsted-Lowry acid or base, using melting point analysis to determine the purity of a compound, and calculating percent yield. Table 1. Table of Reagents Material Hydrochloric

Molecular Weight(g/mol) 36.46

Density(g/mL) 1.2

Boiling Point(°C) -85.05

Melting Point(°C) -114.2

Acid Sodium Hydroxide Diethyl Ether Hexane Ethyl 4 Nitrobenzoate Ethyl 4 Aminobenzoate Sodium Sulfate

39.997

1.515

1388

318

74.12 86.18 181.15

0.706 0.659 1.255

34.6 68.73 279

-116.3 -96 76-80

165.19

1.168

187-189

88-92

142.04

2.66

1429

884

Experimental The first part of this experiment is a liquid-liquid extraction using two dyes, Sudan Orange and Sudan Blue. The relative solubility of each dye is tested in hexane and aqueous solutions containing HCl, NaOH, or distilled water. Based on the relative solubility of the two dyes in these three solutions, an aqueous solution is chosen to be used in extraction of a mixture of the two dyes. NaOH was selected to perform the extraction based on the relative solubility results. 15 mL of the dye mixture is placed into a separatory funnel. 10 mL of the NaOH is added and the funnel is inverted several times to mix the two solutions. The pressure is released periodically between inversions. After inversion, the funnel is placed onto a ring stand while the layers separate. The bottom layer is collected in a beaker before the extraction process is repeated two more times to ensure complete separation of the two layers. The mixture used in this experiment contained a base and a neutral organic substance. Therefore, HCl is used to extract the mixture three times. 15 mL of the mixture is added to the separatory funnel and it is extracted three times with 10 mL of HCl. After the two layers are separated, NaOH is added to the aqueous solution to return the original base to a neutral charge and cause it to fall out of the solution. The solid is separated from the rest of the solution using vacuum filtration and then weighed to calculate percent yield. Because there is a small amount of water left over in the organic layer, it is ‘dried’ by adding sodium sulfate until the solid moves freely around the bottom of the flask. After the solution is dried, it is heated to create a solid. A hot water bath is heated to 40-50 degrees and the organic solution is put into the water bath to heat. The solution is emptied from the flask containing the sodium sulfate before heating. Once a solid is obtained, it is weighed to record percent yield. The melting points of both solids are recorded and compared to known melting points to assess purity. Results Acid Base Extraction of a Dye: When the mixture of the two dyes was separated, the top layer became increasingly bluer with each extraction. The bottom layer turned orange after each extraction, becoming less saturated in

hue as more and more of the solution was extracted. The final result was a bright blue solution left in the funnel with an orange solution in the flask below. Table 2. Relative Solubility of Dyes Sudan Orange Sudan Blue

 

1 M HCl Not soluble, dye on top layer Not soluble, dye on top layer

Distilled Water Not soluble, dye on top layer Not soluble, dye on top layer

1 M NaOH Soluble, no distinct layers Not soluble, dye on top layer

Equation 1: Percent Yield= (Actual Yield/Theoretical Yield)*100% Example Calculation= (.103g/.236g)*100 =43.6%

Percent Yield Calculations:

0.13 g/0.149 g *100%=87.2% yield aqueous layer 0.099 g/0.146 g *100%=67.8% yield organic layer Acid Base Extraction of a Mixture: After the aqueous solution was separated from the organic solution and both were weighed, the percent yield was calculated for each of them.  

Equation 1: Percent Yield= (Actual Yield/Theoretical Yield)*100% Example Calculation= (.103g/.236g)*100 =43.6%

Eq. 1

Percent Yield Calculations:

The theoretical yield for the aqueous layer was calculated to be 0.149 g, and the theoretical yield for the organic layer was calculated to be 0.146 g. The actual yield of the aqueous layer was 0.13 g, so using equation 1, the percent yield was calculated to be 87.2%. The actual yield for the organic layer was 0.099 g, so the percent yield was calculated to be 67.8%. The melting points of both solids were also taken to assess purity of the samples. The melting point of the aqueous layer was found to be about 90°C and the melting point of the organic layer was found to be about 57°C.

Discussion

Week 1: Image 1: Sudan Orange

Image 2: Sudan Blue

Image 3: Hexane

Week 2: Image 4: Mixture 1

Image 5: Conjugate Acid

In the first part of the experiment two dyes were separated using extraction. Based on the second part of the experiment where an acid was used to add charge to a base and the fact that the Sudan Orange did not separate when added to NaOH, it can be inferred

that Sudan Orange is acidic and Sudan Blue is neutral. That is why NaOH was selected to perform the extraction, so that it would deprotonate the acid. For the second part of the experiment, the percent yield of the aqueous layer was higher than the percent yield of the organic layer. In the experimental results on the Canvas course, the percent yield was higher for the organic layer than it was for the aqueous layer.2 This discrepancy could be caused by an error at some point in the experiment. One point of difficulty in this experiment was separating the organic and aqueous layers during extraction. Because there was only a slight color difference, it was difficult to tell the layers apart. This problem could be solved by adding a dye to one of the layers to make it easier to distinguish them. The melting points that we gathered were close to the established melting points of the two solids, indicating that the samples were mostly pure. Because the yield of the organic layer was lower than the yield of the aqueous layer, there may have been problems during the drying and evaporating process. Some solution may have been lost when transferring from the flask with the sodium sulfate to the one that was heated. A more precise way of separating the crystals and the solution could have ensured that no solution was lost. Conclusion This experiment showed the use of acid-base reactions in extractions of mixed solutions. By making one part of the solution polar, it was able to be separated into different layers. While the techniques used in this experiment were relatively simple, there are several errors that could be made. Using the incorrect solvent to protonate or deprotonate the acid or base could ruin the whole experiment. When allowing the bottom layer to escape from the separation funnel, it is possible to let out too much or too little solution, either keeping some of the bottom layer in the funnel with the organic layer or allowing some of the organic layer into the flask with the aqueous layer. If too much acid or base is added when neutralizing the aqueous solution, it could skew the results as well. Errors during filtration, such as not allowing the solid to dry completely, could also impact the calculated percent yield. Despite these possible errors, if care is taken while performing this experiment it can be a relatively quick and easy way of separating polar and nonpolar solutions.

References

(1) https://uab.instructure.com/courses/1532840/files/64421595?module_item_id=15 818748 (accessed Oct 21, 2020). (2) Libretexts. Acid-Base Extraction https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Demos_Techniques_ and_Experiments/General_Lab_Techniques/Acid-Base_Extraction (accessed Oct 21, 2020) (3) Libretexts. Liquid-Liquid Extraction https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Demos_Techniques_and_E xperiments/General_Lab_Techniques/Liquid-Liquid_Extraction (accessed Oct 22, 2020)....