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Chapter 13: Problem 21

The solubility of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water is \(208 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water at \(15^{\circ} \mathrm{C}\). A solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water at \(35^{\circ} \mathrm{C}\) is formed by dissolving \(324 \mathrm{~g}\) in \(100 \mathrm{~g}\) water. When this solution is slowly cooled to \(15^{\circ} \mathrm{C}\), no precipitate forms. (a) What term describes this solution? (b) What action might you take to initiate crystallization? Use molecular- level processes to explain how your suggested procedure works.

Short Answer

Expert verified
(a) The solution is described as supersaturated, as there is more solute present than the solubility allows at \(15^{\circ} \mathrm{C}\). (b) To initiate crystallization, one can introduce a seed crystal, dust, or scratch the container's sides to provide nucleation sites for the crystals. This allows excess solute molecules to align and form a crystalline structure, lowering the overall energy of the system and resulting in crystallization.

Step by step solution

01

To identify the term describing this solution, we must compare the amount of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) dissolved in water at the two different temperatures. At \(15 ^{\circ} \mathrm{C}\), the solubility is \(208 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water. However, at \(35 ^{\circ} \mathrm{C}\), there are \(324 \mathrm{~g}\) of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) dissolved in \(100 \mathrm{~g}\) of water. When this warmer solution is cooled to \(15 ^{\circ} \mathrm{C}\), there is no precipitate formed despite the solubility at this temperature being lower (\((208 \mathrm{~g})\) than the amount of solute present (\((324 \mathrm{~g})\)). This solution is said to be supersaturated since there is more solute present than the solubility allows at the given temperature. #b. Suggesting an action to initiate crystallization#

One action that can be taken to initiate crystallization in a supersaturated solution is to introduce a seed crystal which is a small crystal of the solute that has the same structure as the desired crystal. Alternatively, one could also introduce dust or scratch the container's sides, which can act as a nucleation site for the crystals. These actions will initiate the process of crystallization because the supersaturated solution is unstable. The addition of a seed crystal or the presence of irregularities on the container's inner surface provides a point for the excess solute molecules to latch onto, allowing them to align themselves and form the crystalline structure. This new arrangement will lower the overall energy of the system and result in the crystallization of the excess solute.

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Key Concepts

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

Solubility
Solubility refers to how much of a solute can be dissolved in a solvent at a specific temperature to form a stable solution. It is a measure of the maximum amount of solute that can be dissolved before the solution becomes saturated. For example, the solubility of Cr(NO\(_3\))\(_3\)路9H\(_2\)O in water at 15掳C is 208 g per 100 g of water. This means only up to 208 g of this compound can dissolve in 100 g of water at this temperature.
Understanding solubility is essential because it varies with temperature. Typically, solubility increases as the temperature rises, allowing more solute to dissolve. However, when a solution exceeds its solubility limit upon cooling, it may still remain stable temporarily, creating a supersaturated solution. This state can hold more solute than it normally would at the cooler temperature.
Crystallization
Crystallization is the process where a solution forms solid crystals from the solute. This process can occur when a supersaturated solution is disturbed.
Crystallization begins when solute particles come together to form a nucleus, which grows to form a crystal. The process continues as more solute molecules attach to the forming crystal. This can be induced by adding a seed crystal or by other actions that disturb the solution, like scratching the container. These actions help initiate crystallization by providing points where solute molecules can start organizing into a solid crystal structure.
Nucleation Sites
Nucleation sites are specific spots where crystallization begins. They are crucial for transforming a supersaturated solution into crystals.
These sites can be introduced deliberately by adding a seed crystal, or they might occur naturally from imperfections or dust present in the solution. When you add something like a seed crystal, it acts as a focal point where the excess solute molecules start gathering.
Once gathered, these molecules form an organized pattern, transforming from a disordered solution into an ordered crystal. By decreasing the disorder in the solution, nucleation lowers the energy, encouraging crystallization.
Molecular-Level Processes
At a molecular level, crystallization involves solute molecules leaving the solution phase and joining a solid phase in an orderly fashion. When a supersaturated solution is ready to crystallize, its molecular structure is unstable with more solute than the usual saturation point.
The molecules in a supersaturated solution are constantly moving. They occasionally collide and stick to each other, forming tiny aggregates. If these aggregates house enough molecules, they act as nuclei, prompting further molecules to join and form a crystal.
  • These processes rely heavily on temperature, concentration, and the presence of nucleation sites.
  • Energy dynamics are central here; a stable, lower-energy arrangement is preferred, which is why crystals tend to form from a supersaturated solution.
By understanding these processes, we can control and optimize the conditions for efficient crystallization.

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Most popular questions from this chapter

Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at \(25^{\circ} \mathrm{C}\), and their solubilities in water at \(25^{\circ} \mathrm{C}\) and 1 atm fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution. (b) Explain why the molarity of each of the solutions should be very close numerically to the molality. (c) Based on their molecular structures, account for the differences in solubility of the four fluorocarbons. (d) Calculate the Henry's law constant at \(25^{\circ} \mathrm{C}\) for \(\mathrm{CHClF}_{2}\), and compare its magnitude to that for \(\mathrm{N}_{2}\left(6.8 \times 10^{-4} \mathrm{~mol} / \mathrm{L}-\mathrm{atm}\right) .\) Can you account for the dif- ference in magnitude?

A solution is made containing \(14.6 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in \(184 \mathrm{~g}\) \(\mathrm{H}_{2} \mathrm{O}\). Calculate (a) the mole fraction of \(\mathrm{CH}_{3} \mathrm{OH}\), (b) the mass percent of \(\mathrm{CH}_{3} \mathrm{OH}\), (c) the molality of \(\mathrm{CH}_{3} \mathrm{OH}\).

(a) What is the mass percentage of iodine \(\left(\mathrm{I}_{2}\right)\) in a solution containing \(0.035 \mathrm{~mol} \mathrm{I}_{2}\) in \(115 \mathrm{~g}\) of \(\mathrm{CCl}_{4} ?\) (b) Seawater contains \(0.0079 \mathrm{~g} \mathrm{Sr}^{2+}\) per kilogram of water. What is the concentration of \(\mathrm{Sr}^{2+}\) measured in ppm?

A "canned heat" product used to warm chafing dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin that has an average formula of \(\mathrm{C}_{24} \mathrm{H}_{5}\). What mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) should be added to \(620 \mathrm{~kg}\) of the paraffin in formulating the mixture if the vapor pressure of ethanol at \(35^{\circ} \mathrm{C}\) over the mixture is to be 8 torr? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.

When \(10.0 \mathrm{~g}\) of mercuric nitrate, \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\), is dissolved in \(1.00 \mathrm{~kg}\) of water, the freezing point of the solution is \(-0.162^{\circ} \mathrm{C}\). When \(10.0 \mathrm{~g}\) of mercuric chloride \(\left(\mathrm{HgCl}_{2}\right)\) is dissolved in \(1.00 \mathrm{~kg}\) of water, the solution freezes at \(-0.0685^{\circ} \mathrm{C}\). Use these data to determine which is the stronger electrolyte, \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) or \(\mathrm{HgCl}_{2}\).
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