Solvent Extraction report, Lab Reports of Chemical Separation Processes

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Solvent Extraction Report

• Introduction
• Materials and methods
• Results and discussion
• Conclusion
• References

The classical way of performing LLE is the separation funnel extraction (batch process). However, some LLE extractors or steam distillers are also available, presenting a continuous process. The industry usually prefers continuous operations: it is generally simpler to control automatically, and it makes better use of labor (Kislik 2012). To obtain a large transfer area between two phases dispersion of one of the phases into drops have to be made during the mixing. For this purpose, a construction of type of a gravitational column or a mechanically mixing apparatus is usually used. The last one comprises some means for mixing the two phases.

2.Materials and methods

2.1 Water-Acetone-Toluene 2.1.1 Lab scale extraction A feed solution containing 0,973 g/mL of acetone dissolved in water was mixed with a toluene and put into a separation funnel. The mixture was shaken for 30 minutes at T = 20 ± 2°C. After phase separation, the extract was decanted, weighted and the density was measured via an aerometer. 2.2.2 Multi stage semi technical extraction A feed solution with the same acetone concentration of 0,973 g/mL dissolved in water was also mixed with a toluene. However, in this experiment set-up (Fig. 2) continuous automatic measurements of mass flows and densities of interest can be taken. The semi-technical extractor was operating for approximately 30 minutes also at T = 20 ± 2°C. A density of the obtained extract was measured both automatically (online value can be noted from the control panel) and offline. For the offline measurement a sample of approximately 70 ml organic phase was taken via the valve HV6 to measure its’ density manually, using a set of aerometers.

Figure 2. Experimental scheme of the semi-technical extractor Based on the densities, the molar fraction of acetone can be calculated, using equation 1: 𝑥 (1). 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 = 15, 968ρ 2 − 33, 254ρ + 17, 299 And for toluene, equation 2 can be used 𝑥 (2) 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 = 76, 718ρ 2 − 144, 22ρ + 42, 352 2.2 Metal extraction of Ni and Cu The extraction of metals was performed by mixing equal volumes of aqueous and organic phases (10 mL each) in a separatory funnel. The aqueous phase (feed) contained 498, mg/L of Co and 340,73 mg/L of Ni dissolved in a chloride medium, while the organic phase of Cyanex 272 acted as an extractant in the solution and was observed in various concentrations. 8 samples were made with 2 control results for each concentration of Cyanex 272: 0,2 mol/L; 0,3 mol/L; 0,4 mol/L; 0,5 mol/L. Our group used the 0,3 mol/L concentration as a reference point for the following calculations. The process of shaking two phases in the separatory funnel was carried out for 30 minutes with following operating parameters: T = 20 ± 2°C, 𝝎 = 180 rpm. Once both phases were separated, the aqueous phase was taken out to analyse the resulting Co and Ni concentrations. The mass balance equation was then used to calculate the metal concentration in the organic phase.

𝑥 , whereas =0,854 g/mL 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 = 76, 718ρ 2 − 144, 22ρ + 42, 352 ρ 𝑥 (^) 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 = 0, 76 Remaining part of solution is acetone, therefore 𝑥 𝑎𝑐𝑒𝑡𝑜𝑛𝑒

𝑡𝑜𝑙𝑢𝑒𝑛𝑒

, therefore 𝑤 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = 0, 17 In order to find mass of water feed, 𝑚 𝑓𝑒𝑒𝑑 = ρ 𝑤𝑎𝑡𝑒𝑟

𝑓𝑒𝑒𝑑

𝑓𝑒𝑒𝑑

In order to find mass of extract, 𝑚 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = ρ (^) 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 * 𝑉 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = 868 * 0, 08 (𝑉 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 𝑡𝑎𝑘𝑒𝑛 𝑎𝑠 0, 08 𝐿) 𝑚 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

In order to find mass of acetone feed, 𝑚 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑 = 𝑤 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑 * 𝑚 (^) 𝑓𝑒𝑒𝑑 = 0, 1711946268 * 119, 76 𝑚 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑 = 20, 50 𝑔 In order to find mass of acetone extract, 𝑚 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

Therefore, 𝑚 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑

3.1.2 Method A Replicate 2

Applying in the equation 1 we can obtain 𝑥 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒. 𝑥 , whereas =0,973 g/mL. 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 = 15, 968ρ 2 − 33, 254ρ + 17, 299 ρ 𝑥 𝑎𝑐𝑒𝑡𝑜𝑛𝑒

Remaining part of solution is water, therefore 𝑥 𝑤𝑎𝑡𝑒𝑟

𝑎𝑐𝑒𝑡𝑜𝑛𝑒

, therefore 𝑤 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑 = 0, 17 Applying in the equation 2 we can obtain 𝑥 (^) 𝑡𝑜𝑙𝑢𝑒𝑛𝑒. 𝑥 (^) 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 = 76, 718ρ , whereas =0,855 g/mL 2 − 144, 22ρ + 42, 352 ρ 𝑥 𝑡𝑜𝑙𝑢𝑒𝑛𝑒

Remaining part of solution is acetone, therefore 𝑥 𝑎𝑐𝑒𝑡𝑜𝑛𝑒

𝑡𝑜𝑙𝑢𝑒𝑛𝑒

, therefore 𝑤 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = 0, 15 𝑚 𝑤𝑎𝑡𝑒𝑟

In order to find mass of acetone extract, 𝑚 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑚 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = ρ (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 * 𝑉 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = 0, 852 * 2000 (𝑉 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 𝑡𝑎𝑘𝑒𝑛 𝑎𝑠 2000 𝑚𝐿/ℎ) 𝑚 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = 1704 𝑔/ℎ In order to find mass flow of acetone feed, 𝑚 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑

𝑓𝑒𝑒𝑑

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑

In order to find mass flow of acetone extract, 𝑚 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑒𝑥𝑡𝑟𝑎𝑐𝑡

Therefore, 𝑚 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑

3.1.4 Method B Offline 2 Applying in the equation 1 we can obtain 𝑥 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒. 𝑥 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 = 15, 968ρ , whereas =0,973 g/mL. 2 − 33, 254ρ + 17, 299 ρ 𝑥 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 = 0, 06 Remaining part of solution is water, therefore 𝑥 (^) 𝑤𝑎𝑡𝑒𝑟 = 1 − 𝑥 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 = 1 − 0, 060226672 = 0, 94 , therefore 𝑤 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑

Applying in the equation 2 we can obtain 𝑥 (^) 𝑡𝑜𝑙𝑢𝑒𝑛𝑒. 𝑥 (^) 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 = 76, 718ρ , whereas =0,8502 g/mL 2 − 144, 22ρ + 42, 352 ρ 𝑥 (^) 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 = 0, 7

, therefore 𝑤 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = 0, 22 In order to find mass flow of water feed, 𝑚 𝑓𝑒𝑒𝑑 = ρ 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑓𝑒𝑒𝑑

𝑓𝑒𝑒𝑑

In order to find mass flow of extract, 𝑚 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = ρ (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 * 𝑉 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 = 0, 8502 * 2000 (𝑉 (^) 𝑒𝑥𝑡𝑟𝑎𝑐𝑡 𝑡𝑎𝑘𝑒𝑛 𝑎𝑠 2000 𝑚𝐿/ℎ) 𝑚 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

In order to find mass flow of acetone feed, 𝑚 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑 = 𝑤 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑 * 𝑚 (^) 𝑓𝑒𝑒𝑑 = 0, 17 * 2919 𝑚 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑 = 499, 72 𝑔/ℎ In order to find mass flow of acetone extract, 𝑚 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡

𝑒𝑥𝑡𝑟𝑎𝑐𝑡

Therefore, 𝑚 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑒𝑥𝑡𝑟𝑎𝑐𝑡/𝑚 (^) 𝑎𝑐𝑒𝑡𝑜𝑛𝑒 𝑓𝑒𝑒𝑑 = 0, 7319 = 73, 19% Summary of results for Water-Acetone-Toluene experiment is given below. Table 3. Water-Acetone-Toluene Experiment Results Streams Density 20 ºC (g/ml) Molar fraction xacetone Mass fraction wacetone Feed 0,973 0,060 0, Method A Extract 1 0,854 0,240 0,

Ni 57,86^ 16,98^ 0, Figure 3. Extraction Percentage of Each Group This shows that as the concentration of Cyanex 272 was increased so was the % E for both elements. Because there were more “available” extractant molecules in the organic phase to interact with the metals. But always Co had the higher values of extraction, this behavior for the extractant can also be noticed in other works (Scal, Seruff and Vera, 2020). Cyanex 272 extracts preferably metals with higher molar mass.

Figure 4. Beta Co/Ni Values of Each Group In order to compare the efficiency of Cyanex 272 concentration, Beta Co/Ni values were analysed from the graph above. It can be noticed from the values given in the graph above that 0,2 mol/L of Cyanex 272 (GI and GII) resulted in considerably higher Beta Co/Ni values (28 and 31 approximately) than other given concentrations of Cyanex 272. Therefore, 0,2 mol/L of Cyanex 272 concentration is more efficient than other given concentrations because Beta expresses how well the two elements were separated. To reach the maximum Co extraction with Vorg/Vaq = 1,0 would be very unlikely, as shown in graph below, using McCabe-Thiele method (Kislik, 2012). The operation line would overlap the curve with the equilibrium points (obtained with the different concentrations used by the groups). This prevents us from estimating how many steps would be necessary to separate the metals.

Changing the ratio A/O allowed us to estimate the steps because the operation line was under the equilibrium curve.

4.Conclusion

For the first part of the practical (acetone-water-toluene) it could be noticed that method A (one step lab scale extraction) led to a lower extraction of acetone when compared to method B (multi stage semi technical extraction). The results suggest that the lab scale didn’t provide as much diffusion between aqueous and organic phase as the multi stage did. As we had a maximum mass fraction of acetone 0.17 for method A and 0.21 for method B, it means that the contact between the phases was more efficient in B, allowing the acetone molecules to migrate to the organic phase, therefore leading to a higher mass fraction in the extract (organic phase pos extraction). Regarding the metal extraction, the results suggest that the extractant Cyanex 272 preferably extracts the heaviest element, in this case, Cobalt. Besides the higher extraction of Co, the separation factor was inversely proportional to the increase of Cyanex 272 concentration, thus the highest value of this separation factor (31,38) was obtained with the lowest value of Cyanex concentration (0.2 mol/L). This means that high concentrations of the extractant may lead to high extraction percentages but for both elements (Co and Ni) not separating them very well.

5.References

Kislik, V. S. (2012b). Engineering Development of Solvent Extraction Processes. Solvent Extraction, 157–184. https://doi.org/10.1016/b978-0-444-53778-2.10004- Moldoveanu, S., & David, V. (2015). Solvent Extraction. In Modern Sample Preparation for Chromatography. https://doi.org/10.1016/b978-0-444-54319-6.00006- Scal, M L; Seruff L A; Vera Y M. Study of the separation of didymium from lanthanum using liquid-liquid extraction: Comparison between saponification of the extractant and use of lactic acid. Minerals Engineering, V. 148, 2020. ZHANG, J; ZHAO, B. Separation hydrometallurgy of rare earth elements. Switzerland: Springer International Publishing, 2016.