Question

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 \(\mathrm{g}\) of water. When this solution is slowly cooled to \(15^{\circ} \mathrm{C},\) no precipitate forms. (a) Is the solution that has cooled down to \(15^{\circ}\) Cunsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

Short answer

(a) The solution at 15°C is supersaturated. (b) The process of nucleation occurred when the glass was scratched with the spatula, which led to the appearance of crystals. (c) At equilibrium, 116g of crystals are expected to form.

Step-by-step solution

Determine the mass of solute in a saturated solution at 15°C

To do this, we will use the given solubility of the salt: 208g per 100g of water. So, in a saturated solution at 15°C, there will be 208g of solute dissolved in 100g of water.

Analyze the solution at 35°C

We are given that at 35°C, 324g of salt is dissolved in 100g of water. Meanwhile, the solubility at 15°C is 208g of solute per 100g of water.

Determine the state of the solution at 15°C

As the solution cools down to 15°C, the solubility decreases to 208g/100g of water. The salt remains dissolved in the solution, so it is now a supersaturated solution because it contains more solute than the saturation point at 15°C. Answer (a): The solution at 15°C is supersaturated.

Scratch the glass and explain crystals appearing process

When the glass is scratched with the spatula, it introduces small disturbances to the supersaturated solution, providing nucleation sites where the excess solute can attach to and start crystallizing. This is known as nucleation, which is the initial event that leads to crystal growth when the conditions of a supersaturated solution are altered. Answer (b): The process of nucleation occurred when the glass was scratched with the spatula, which led to the appearance of crystals.

Calculate the mass of crystals expected at equilibrium

In a saturated solution, there would be 208g of solute in 100g of water at 15°C. Since the solution contains 324g of solute, the difference between these values represents the mass of crystals expected to form at equilibrium: Mass of crystals = 324g (initial) - 208g (saturated amount at 15°C) = 116g Answer (c): At equilibrium, 116g of crystals are expected to form.

Key concepts

Supersaturated Solutions

When a solution contains more solute than it can usually dissolve at a given temperature, it is called supersaturated. In the exercise, as the solution of chromium(III) nitrate from 35°C is cooled to 15°C, it remains clear with all 324g of solute still dissolved in 100g of water. This situation is interesting because at 15°C, the maximum solubility should only allow for 208g to dissolve. Thus, more solute is dissolved than what is theoretically possible at that temperature. This is a classic example of a supersaturated solution.

Supersaturated solutions are delicate. They often lack stable conditions and are in search of a way to relieve the 'excess' of solute in them. This excess makes them prone to crystal formation, which can happen spontaneously or can be triggered by minor disturbances. Due to this sensitivity, they are often used in various scientific experiments to study solubility and crystallization processes.

Nucleation Process

The appearance of crystals from a supersaturated solution is involuntarily fascinating and is guided by the nucleation process. Nucleation is the initial step before crystals can grow and develop further in size. It occurs when the excess solute begins to gather at specific points, called nucleation sites. These sites can be introduced naturally or can be triggered by external changes in the environment of the solution.

In the given exercise, when a spatula scratched the glass container holding the supersaturated solution, microscopic disturbances were created. These disturbances served as nucleation sites. This prompted the extra solute in the solution to begin accumulating in tiny clusters, eventually forming crystals. Without nucleation, even though a solution might be supersaturated, solute particles need a starter point to transition into a solid state. Hence, nucleation is crucial for crystal formation.

Crystal Formation

Crystals are not only beautiful but are also a result of precise and specific molecular arrangement. When a supersaturated solution undergoes nucleation, its solute molecules start to bond in a repeating, orderly fashion, forming crystal structures. This process can be fascinating to observe as it represents a shift from a random state to an organized state.

In the exercise, after the nucleation was triggered by scratching the glass, 116 grams of crystals were expected to precipitate. This is a tangible outcome of the solution aiming to attain equilibrium. As the saturated solubility limit at 15°C is only 208g, the 116g of excess solute needed to find a stable phase, which is crystal form. The resulting crystals reflect the excess solute leaving the supersaturated solution to restore a state of equilibrium.

• Supersaturation causes potential for crystals to form due to excess solute.
• Nucleation triggers crystal growth by providing gathering points for solute molecules.
• Final crystals symbolize the natural balance between dissolved solute and solid form.

Understanding this transformation from a supersaturated solution to a crystal structure helps to appreciate the forces and conditions that govern crystallization.