/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 Various amino acids have utility... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Various amino acids have utility as food additives and in medical applications. They are often synthesized by fermentation using a specific microorganism to convert a substrate (e.g., a sugar) into the desired product. Small quantities of other species also may be formed and must be removed to meet product specifications. For example, isoleucine (Ile), which has a molecular weight of \(131.2,\) is an essential amino acid \(^{16}\) produced by fermentation, and other amino acids such as leucine and valine also are found in the fermentation broth. The broth is subjected to several processing steps to remove these and other impurities, but final processing by crystallization is required to meet stringent specifications on purity. The strategy is to crystallize the hydrated acid form of Ile (Ile. \(\mathrm{HCl} \cdot \mathrm{H}_{2} \mathrm{O}\) ), whose crystals exclude other amino acids, and then to redissolve, neutralize, and crystallize the final Ile product. In a batch process designed to manufacture \(2500 \mathrm{kg}\) of Ile per batch, an aqueous feed solution containing 35 g Ile/dL and much lower concentrations of leucine and valine is fed to the final purification stages. The pH of the solution is 1.1 and its specific gravity is 1.02. The solution is heated to \(60^{\circ} \mathrm{C}\) and 35-wt\% HCl solution is added in a ratio of 0.4 kg per kg of feed. The addition of HCl causes the formation of crystals of Ile\cdotHCl\cdot \(\mathrm{H}_{2} \mathrm{O},\) and the production of these crystals is further increased by slowly lowering the temperature to \(20^{\circ} \mathrm{C}\). At the final crystallizer conditions the Ile solubility is \(5 \mathrm{g}\) Ile/ \(100 \mathrm{g}\) solution. The resulting slurry is sent to a centrifuge where the crystals are separated from the liquid solution and the crystal cake is washed with water. The solids leaving the centrifuge contain \(12 \%\) free water (i.e., not part of the crystal structure) and \(88 \%\) pure crystals of Ile\(\cdot \mathrm{HCl} \cdot \mathrm{H}_{2} \mathrm{O}\). \(\mathrm{H}_{2} \mathrm{O}\).The washed crystals "water to form a solution that is 4.0 g Ile/dL with gravity of 1.1. The solution is sent to an ion exchange unit where HCl is removed. Upon leaving the ion exchange unit the solution has a pH of about \(5.5 .\) It is sent to a second crystallizer where the temperature is gradually reduced to \(10^{\circ} \mathrm{C}\) and the Ile solubility is \(3.4 \mathrm{g} \mathrm{Ile} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). The crystals are separated from the slurry by centrifugation, washed with pure water, and sent to a dryer for final processing. (a) Construct a labeled flowchart for the process. (b) Choosing a basis of 1 kg of feed solution, estimate (i) the mass of HCl solution added to the system, (ii) the water added to redissolve the Ile.HCI. \(\mathrm{H}_{2} \mathrm{O}\) crystals, (iii) the mass of \(\mathrm{HCl}\) removed in the ion exchange unit, and (iv) the mass of final Ile product. (c) Scale the quantities calculated in Part (b) to the production rate of 2500 kg Ile/batch. (d) Estimate the active volume (in liters) of each of the crystallizers. (e) Amino acids are amphoteric, which means they can either donate or accept a proton \(\left(\mathrm{H}^{+}\right) .\) At low pH they tend to accept a proton and become acidic while at high pH they tend to donate a proton and become basic. They also are known as zwitterions because their ends are oppositely charged, even though the overall molecule is neutral. Isoleucine is reported to have an isoelectric point (pI) of 6.02 and \(\mathrm{pK}_{\mathrm{a}}\) values of 2.36 and \(9.60 .\) Look up the meaning of these terms and prepare a plot showing how these values are used in plotting the distribution of Ile between acid, zwitterionic (neutral), and basic forms as a function of pH. Explain why such a distribution is important in carrying out the separations described in the process.

Short Answer

Expert verified
The mass of the HCl solution added is 0.4 kg, the mass of water used to redissolve the crystals is 1060 g, the mass of HCl removed in the ion exchange step will be equal to the moles of Ile present in the crystal form. The mass of the final Ile product can be calculated as 3.4% of the total mass of the final solution. All these values will need to be scaled as per the batch size. The volumes of the crystallizers could be estimated by assuming full occupation by the solutions in them. The pH in every step plays a key role in facilitating the separation of Ile from other amino acids and impurities due to their zwitterionic nature and specific pKa values.

Step by step solution

01

Set Up the Flowchart for the Process

You'll need to draw a flowchart that illustrates the process of manufacturing isoleucine. It should include the steps of fermentation, the addition of HCl, crystallization, re-dissolution, ion exchange, re-crystallization, and final processing. For each step, label the inputs, outputs, and conditions (e.g. pH, temperature).
02

Calculate the Mass of the HCl Solution Added

This can be found from the given ratio of 0.4 kg of HCl solution per kg of feed. Since the basis is 1 kg of feed, the mass of HCl solution added is 0.4 kg.
03

Find the Water Added to Redissolve the Crystals

The mass of water added can be calculated from the declared solubility of the crystals in water, which is 4 g Ile/dL. The resulting solution has a specific gravity of 1.1, meaning its density is \(1.1 \times 1 \mathrm{g/mL} = 1.1 \mathrm{g/mL}\). A 1 L or 10 dL of solution will therefore have a mass of \(1.1 \times 1000 \mathrm{g} = 1100\mathrm{g}\). With 4 g of Ile per dL of solution, there are \(4 \times 10 \mathrm{g} = 40 \mathrm{g}\) of Ile in 1L of solution, so the remaining mass, \(1100 \mathrm{g} - 40 \mathrm{g} = 1060 \mathrm{g}\) will be the water added.
04

Calculate the Mass of HCl Removed

This can be found through the molar ratio of Ile to HCl in the crystal form. Since they form a 1:1 ratio, the moles of Ile present, calculated from the mass and molar mass, will also be the moles of HCl present. The resulting mass of HCl can then be calculated from these moles and the molar mass of HCl.
05

Calculate the Mass of Final Ile Product

The final product concentration is given as 3.4 g Ile per 100 g of solution, and the mass of final product can therefore be calculated as a percentage of the total mass of solution at this step.
06

Scale Calculations to Batch Size

This requires simple multiplication of the values obtained in steps 2-5 by the scaling factor (the ratio of batch size to basis size).
07

Estimate the Volume of the Crystallizers

Assuming the crystallizers are completely filled with solution during operation, their volume could be estimated from the mass and density of the solution at the relevant step.
08

Explain the Importance of pH in the Separation Process

This will involve discussing how amino acids act as buffers and how the changes in pH throughout the process facilitate the separation of Ile from other amino acids and impurities, drawing upon the knowledge of pKa values and zwitterionic nature of amino acids.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amino Acid Crystallization
Amino acid crystallization is a key step in ensuring the purity of the desired product, such as isoleucine. In the crystallization process, amino acids are precipitated out of a solution in a crystalline form. This is achieved by manipulating conditions such as temperature and pH level.
This step is critical because it separates the target amino acid from impurities and by-products, making it suitable for use in food and medical applications. The crystallization process typically involves:
  • Adjusting the temperature: Lowering the temperature decreases the solubility of amino acids, promoting crystallization.
  • pH adjustment: Adjusting the pH can change the charge and solubility of the amino acids, influencing their crystallization behavior.
  • Use of solvents: Sometimes, the addition of specific solvents can help in selectively precipitating the desired amino acid.
Crystallization is often followed by filtration or centrifugation to separate solid crystals from the liquid, enhancing the purity of the final product.
Fermentation Process
The fermentation process is essential in the production of amino acids like isoleucine. This biochemical process involves the use of microorganisms to convert sugar substrates into desired amino acids.
During fermentation, specific strains of bacteria or yeast utilize substrates, such as glucose, to produce amino acids as metabolic by-products. Key factors in the fermentation process include:
  • Microbial selection: Using specific microorganisms that efficiently produce the desired amino acid.
  • Optimized conditions: Controlling temperature, pH, and nutrient availability to maximize yield.
  • Sterile environment: Ensuring the fermentation takes place in a sterile environment to prevent contamination.
Once the fermentation process is complete, the broth containing the amino acids is subjected to purification steps, including crystallization, to isolate and refine the product.
Isoleucine Production
Isoleucine, an essential amino acid, is produced through a meticulous process of fermentation and purification. The production process aims to achieve high purity isoleucine, meeting stringent pharmaceutical and food industry standards.
The production begins with fermentation, where microbes convert sugars into a rich broth of amino acids, including isoleucine. Following this, various purification steps such as crystallization and ion exchange are employed. The critical steps in isoleucine production include:
  • Fermentation: Microorganisms ferment a sugar-rich substrate to produce isoleucine.
  • Crystallization: Through controlled crystallization, isoleucine is separated from other amino acids and impurities.
  • Ion Exchange: This step further purifies the isoleucine by removing unwanted ions from the solution.
  • Final Crystallization: Lowering the temperature to promote further crystallization of pure isoleucine.
These steps ensure the production of high purity isoleucine, suitable for diverse applications, from dietary supplements to pharmaceuticals.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An air conditioner is designed to bring \(10,000 \mathrm{ft}^{3} / \mathrm{min}\) of outside air \(\left(90^{\circ} \mathrm{F}, 29.8 \text { inches } \mathrm{Hg} .88 \%\right.\) relative humidity) to \(40^{\circ} \mathrm{F}\), thereby condensing a portion of the water vapor, and then to reheat the air before releasing it into a room at \(65^{\circ} \mathrm{F}\). Calculate the rate of condensation (gallons \(\mathrm{H}_{2} \mathrm{O} / \mathrm{min}\) ) and the volumetric flow rate of the air delivered to the room. (Suggestion: On the flowchart, treat the coolingcondensation and the reheating as separate process steps.)

The separation of aromatic compounds from paraffins is essential in producing many polyesters that are used in a variety of products. When aromatics and paraffins have the same number of carbon atoms, they often have similar vapor pressures, which makes them difficult to separate by distillation. Extraction is a viable alternative, as illustrated by the following simple system.Sulfolane (an industrial solvent) and octane may be considered completely immiscible. At \(25^{\circ} \mathrm{C},\) the ratio of the mass fraction of xylene in the octane-rich phase to the mass fraction of xylene in the sulfolanerich phase is 0.25. One hundred kg of pure sulfolane are added to 100 kg of a mixture containing 75 wt\% octane and \(25 \%\) xylene, and the resulting system is allowed to equilibrate. How much xylene transfers to the sulfolane phase?

Air containing 20.0 mole \(\%\) water vapor at an initial pressure of 1 atm absolute is cooled in a 1 -liter sealed vessel from \(200^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\).(a) What is the pressure in the vessel at the end of the process? (Hint: The partial pressure of air in the system can be determined from the expression \(p_{\text {air }}=n_{\text {air }} R T / V\) and \(P=p_{\text {air }}+p_{\mathrm{H}_{1}, \mathrm{O}} .\) You may neglect the volume of the liquid water condensed, but you must show that condensation occurs.) (b) What is the mole fraction of water in the gas phase at the end of the process?(c) How much water (grams) condenses?

Ethyl alcohol has a vapor pressure of \(20.0 \mathrm{mm} \mathrm{Hg}\) at \(8.0^{\circ} \mathrm{C}\) and a normal boiling point of \(78.4^{\circ} \mathrm{C}\). Estimate the vapor pressure at \(45^{\circ} \mathrm{C}\) using \((\mathrm{a})\) the Antoine equation; (b) the Clausius-Clapeyron equation and the two given data points; and (c) linear interpolation between the two given points. Taking the first estimate to be correct, calculate the percentage error associated with the second and third estimates.

Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at \(58.3^{\circ} \mathrm{C}\), the height of the mercury in the arm of the manometer connected to the flask is \(747 \mathrm{mm}\) and that in the arm open to the atmosphere is \(52 \mathrm{mm} . \mathrm{At} 110^{\circ} \mathrm{C},\) the mercury level is \(577 \mathrm{mm}\) in the arm connected to the flask and \(222 \mathrm{mm}\) in the other arm. Atmospheric pressure is \(755 \mathrm{mm} \mathrm{Hg}\). (a) Extrapolate the data using the Clausius-Clapeyron equation to estimate the vapor pressure of chlorobenzene at \(130^{\circ} \mathrm{C}\). (b) Air saturated with chlorobenzene at \(130^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) is cooled to \(58.3^{\circ} \mathrm{C}\) at constant pressure. Estimate the percentage of the chlorobenzene originally in the vapor that condenses. (See Example 6.3-2.)(c) Summarize the assumptions you made in doing the calculation of Part (b).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.