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Chapter 6: Problem 81

An aqueous solution of potassium hydroxide (KOH) is fed to an evaporative crystallizer at a rate of 875 kg/h. The crystallizer operates at 10^'C and produces crystals of KOH-2H_O. Water evaporated from the crystallizer flows to a condenser, and the resulting condensate is collected in a tank. During a 30-minute period, 73.8 kg of water is collected. Five-gram samples of the feed to the crystallizer and the liquid removed with the crystals are taken for analysis and subsequently titrated with \(0.85 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) It is found that \(22.4 \mathrm{mL}\) of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is required for the feed and \(26.6 \mathrm{mL}\) is required for the product liquid.(a) What fraction of the KOH in the feed is crystallized? (b) Later you learn that a solution in equilibrium with KOH \(\cdot 2 \mathrm{H}_{2} \mathrm{O}\) crystals at \(10^{\circ} \mathrm{C}\) has a concentration of \(103 \mathrm{kg} \mathrm{KOH} / 100 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} .\) How would this information cause you to reconsider the procedure by which a sample of the mother liquor was obtained? (Hint: Consider removing a slurry sample- -i.e., one containing both solution and KOH \(\cdot 2 \mathrm{H}_{2} \mathrm{O}\) crystals - that is maintained at \(10^{\circ} \mathrm{C},\) but that initially had a solute concentration of \(121 \mathrm{kg} \mathrm{KOH} / 100 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} .\) What would that concentration be after the sample is stored for several hours?)

Short Answer

Expert verified
This problem seems to contain an error as more KOH was found in the product than in the feed which is impossible. An error might occur during the collection of samples or it contains a typographical error.

Step by step solution

01

Understanding the Problem

Here, there is a process where potassium hydroxide (KOH) is fed into a crystallizer to produce crystals of KOH-2H2O. By understanding the weight of water collected and knowing the sample chemical concentrations, the goal is to find what fraction of KOH in the feed is crystallized and understand the effect of solution equilibrium with crystals on the process of sample collection.
02

Calculate Amount of Molecule in Feed

The feed requires 22.4 mL of 0.85 M H2SO4 to titrate, 1 mole H2SO4 neutralizes 2 moles KOH. Therefore, the moles of KOH in the feed is \(0.0224 L\times0.85 mol/L\times2=0.03808 mol\). As the molar mass of KOH is 56.1 g/mol, the mass of KOH in the 5g feed is \(0.03808 mol\times56.1g/mol=2.136 g\).
03

Calculate Amount of Molecule in Product

The product liquid requires 26.6 mL of 0.85 M H2SO4 to titrate, 1 mole H2SO4 neutralizes 2 moles KOH so we can calculate the mass of KOH in the product liquid in the same way. The moles of KOH in the product is \(0.0266 L\times 0.85 moL/L\times 2=0.04526 mol\). Therefore, the mass of KOH in the 5g product liquid is \(0.04526 mol\times 56.1 g/mol=2.537 g\).
04

Calculate Crystallized KOH Fraction

The mass of KOH that crystallizes is the mass of KOH in the feed minus the mass of KOH in the product. This is \(2.136 g - 2.537 g = -0.401 g\). The minus sign indicates that there is more KOH in the product than in the feed which is impossible. Therefore, there might be an error in the problem or the method of analysis, thus no valid crystallized KOH fraction can be calculated in this step.
05

Rechecking the Sampling Process

Regarding the question on equilibrium, the equilibrium concentration of the solution is 103 kg KOH / 100 kg H2O. This tells that the saturated solution at 10 degrees Celsius should not contain more than this concentration. Thus, if a sample was originally taken with concentration higher than this, it would tend to precipitate out or crystallize until reaching equilibrium if its temperature was kept at 10 degrees Celsius. This could cause an error in the original sample analysis, which may explain contradiction in step 4.

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Key Concepts

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

Crystallization
Crystallization is a fundamental process in chemical engineering crucial for purifying substances or preparing solids from a solution. To understand it, think of lowering the solubility of a compound in a solvent, encouraging the formation of solid crystals. This can occur via cooling, evaporating the solvent, or introducing a precipitant.
For instance, in our exercise, potassium hydroxide (KOH) is fed into a crystallizer where water evaporates, which reduces its solubility. This process results in the formation of solid KOH-2H鈧侽 crystals. The crucial aspect is the conditions maintained within the crystallizer, such as temperature and concentration.
  • Cooling the solution, as seen at 10掳C, further promotes crystallization by reducing solubility.
  • Evaporating water effectively increases solute concentration, pushing it past its solubility limit.
As crystallization occurs, sampling must be precise. Consider the equilibrium concentration to ensure samples retain their solute level when solid crystals form. This principle is essential in analyzing and troubleshooting crystallization processes in chemical engineering.
Titration
Titration is a common laboratory technique used to determine the concentration of a given substance in a solution. It involves adding a titrant of known concentration to the solution until the reaction reaches its end point, often indicated by a color change.
In this exercise, sulfuric acid ( H鈧係O鈧�) is used to determine the amount of KOH present in the crystallizer feed and the liquid product. By measuring how much acid is required to neutralize the basic KOH solution, you can calculate its concentration.
  • For example, titrating with 22.4 mL of 0.85 M H鈧係O鈧� helps find the KOH concentration in the feed.
  • Titrating with 26.6 mL of the same acid determines KOH concentration in the product.
This data is then used to calculate the fraction of KOH that has crystallized. It is critical to conduct accurate measurements for precise calculations, particularly essential in process monitoring and quality control in industrial applications.
Equilibrium Concentration
Equilibrium concentration is the stable concentration of solutes in a solution when the rate of dissolution equals the rate of crystallization. At this point, the solution is saturated, meaning no additional solute can dissolve at a given temperature.
In the scenario given, the equilibrium concentration of KOH in solution is 103 kg per 100 kg of H鈧侽 at 10掳C. If the solution initially had a higher concentration than this, crystals would form until the system reached equilibrium.
This principle affects the sampling process. If a sample is taken above equilibrium concentration, over time, it will lose solute to precipitation or crystallization. Hence, understanding equilibrium concentration is important.
  • Ensures that sample analyses reflect actual conditions within a system.
  • Avoids errors introduced by precipitation during sample storage.
Keeping these dynamics in mind aids in taking accurate samples and avoiding misleading conclusions which are vital for maintaining process efficiency and product quality.

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Most popular questions from this chapter

A quantity of methyl acetate is placed in an open, transparent, three-liter flask and boiled long enough to purge all air from the vapor space. The flask is then sealed and allowed to equilibrate at \(30^{\circ} \mathrm{C},\) at which temperature methyl acetate has a vapor pressure of \(269 \mathrm{mm}\) Hg. Visual inspection shows \(10 \mathrm{mL}\) of liquid methyl acetate present.(a) What is the pressure in the flask at equilibrium? Explain your reasoning.(b) What is the total mass (grams) of methyl acetate in the flask? What fraction is in the vapor phase at equilibrium?(c) The above answers would be different if the species in the vessel were ethyl acetate because methyl acetate and ethyl acetate have different vapor pressures. Give a rationale for that difference.

The solubility coefficient of a gas may be defined as the number of cubic centimeters (STP) of the gas that dissolves in \(1 \mathrm{cm}^{3}\) of a solvent under a partial pressure of 1 atm. The solubility coefficient of \(\mathrm{CO}_{2}\) in water at \(20^{\circ} \mathrm{C}\) is \(0.0901 \mathrm{cm}^{3} \mathrm{CO}_{2}(\mathrm{STP}) / \mathrm{cm}^{3} \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\). (a) Calculate the Henry's law constant in atm/mole fraction for \(\mathrm{CO}_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) from the given solubility coefficient. (b) How many grams of \(\mathrm{CO}_{2}\) can be dissolved in a \(12-\mathrm{oz}\) bottle of soda at \(20^{\circ} \mathrm{C}\) if the gas above the soda is pure \(\mathrm{CO}_{2}\) at a gauge pressure of 2.5 atm ( 1 liter \(=33.8\) fluid ounces)? Assume the liquid properties are those of water. (c) What volume would the dissolved \(C O_{2}\) occupy if it were released from solution at body temperature and pressure \(-37^{\circ} \mathrm{C}\) and 1 atm?

The feed to a distillation column (sketched below) is a 45.0 mole\% \(n\) -pentane- 55.0 mole\% n-hexane liquid mixture. The vapor stream leaving the top of the column, which contains 98.0 mole\% pentane and the balance hexane, goes to a total condenser (which means all the vapor is condensed). Half of the liquid condensate is returned to the top of the column as reflux and the rest is withdrawn as overhead product (distillate) at a rate of \(85.0 \mathrm{kmol} / \mathrm{h}\). The distillate contains \(95.0 \%\) of the pentane fed to the column. The liquid stream leaving the bottom of the column goes to a reboiler. Part of the stream is vaporized; the vapor is returned to the bottom of the column as boilup, and the residual liquid is withdrawn as bottoms product.(a) Calculate the molar flow rate of the feed stream and the molar flow rate and composition of the bottoms product stream. (b) Estimate the temperature of the vapor entering the condenser, assuming that it is saturated (at its dew point) at an absolute pressure of 1 atm and that Raoult's law applies to both pentane and hexane. Then estimate the volumetric flow rates of the vapor stream leaving the column and of the liquid distillate product. State any assumptions you make. (c) Estimate the temperature of the reboiler and the composition of the vapor boilup, again assuming operation at 1 atm.(d) Calculate the minimum diameter of the pipe connecting the column and the condenser if the maximum allowable vapor velocity in the pipe is \(10 \mathrm{m} / \mathrm{s}\). Then list all the assumptions underlying the calculation of that number.

A stage of a separation process is defined as an operation in which components of one or more feed streams divide themselves between two phases, and the phases are taken off separately. In an ideal stage or equilibrium stage, the effluent (exit) streams are in equilibrium with each other.Distillation columns often consist of a series of vertically distributed stages. Vapor flows upward and liquid flows downward between adjacent stages; some of the liquid fed to each stage vaporizes,and some of the vapor fed to each stage condenses. A representation of a section of a distillation column is shown below. (See Problem 4.42 for a more realistic representation.) Consider a distillation column operating at 0.4 atm absolute in which benzene and styrene are being separated. A vapor stream containing 65 mole\% benzene and 35 mole\% styrene enters stage 1 at a rate of \(200 \mathrm{mol} / \mathrm{h}\), and liquid containing 55 mole\% benzene and 45 mole\% styrene leaves this stage at a rate of 150 mol/h. You may assume (1) the stages are ideal, (2) Raoult's law can be used to relate the compositions of the streams leaving each stage, and (3) the total vapor and liquid molar flow rates do not change by a significant amount from one stage to the next.(a) How would you expect the mole fraction of benzene in the liquid to vary from one stage to another, beginning with stage 1 and moving up the column? In light of your answer and considering that the pressure remains essentially constant from one stage to another, how would you then expect the temperature to vary at progressively higher stages? Briefly explain. (b) Estimate the temperature at stage 1 and the compositions of the vapor stream leaving this stage and the liquid stream entering it. Then repeat these calculations for stage 2 . (c) Describe how you would calculate the number of ideal stages required to reduce the styrene content of the vapor to less than 5 mole\%.

The constituent partial pressures of a gas in equilibrium with a liquid solution at \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{atm}\) containing \(2 \mathrm{Ib}_{\mathrm{m}} \mathrm{SO}_{2} / 100 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O}\) are \(p_{\mathrm{H}_{2} \mathrm{O}}=31.6 \mathrm{mm} \mathrm{Hg}\) and \(p_{\mathrm{SO}_{2}}=176 \mathrm{mm} \mathrm{Hg} .\) The balance of the gas is air.(a) Calculate the partial pressure of air. If you make any assumptions, state what they are. (b) Suppose the only data available on this system gave \(p_{\mathrm{SO}_{2}}=176 \mathrm{mm}\) Hg, but there was no information given on the equilibrium partial pressure of water. Use Raoult's law to estimate a value for this quantity.Assuming that the value given in the problem statement is correct, what percentage error results from using Raoult's law? (c) The same system was examined in Example \(6.4-1 .\) What percentage errors in the two calculated quantities would result from using Raoult's law for the partial pressure of water?
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