Problem 23 - Chemistry The Central Science In Si Units

Question

The solubility of alum, \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2} \cdot 12 \mathrm{H}_{2} \mathrm{O},\) in water at is \(44 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water at \(50^{\circ} \mathrm{C}\). A solution of alum in water at \(80^{\circ} \mathrm{C}\) is formed by dissolving \(130 \mathrm{~g}\) in \(100 \mathrm{~g}\) of water. When this solution is slowly cooled to \(50^{\circ} \mathrm{C}\), no precipitate forms. (a) Is the solution that has cooled down to \(50^{\circ} \mathrm{C}\) unsaturated, 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?

Step-by-Step Solution

Answer

(a) Supersaturated; (b) Crystallization induced; (c) 86 g of crystals form.

Step 1: Determine solubility at 50°C

The solubility of alum, \(\mathrm{KAl(SO_4)_2 \cdot 12H_2O}\), at \(50^{\circ} \mathrm{C}\) is given as \(44 \mathrm{\, g}\) per \(100 \mathrm{\, g}\) of water. This means at this temperature, \(100 \mathrm{\, g}\) of water can dissolve \(44 \mathrm{\, g}\) of alum to form a saturated solution.

Step 2: Evaluate the solution concentration at 80°C

We have a solution at \(80^{\circ} \mathrm{C}\) containing \(130 \mathrm{\, g}\) of alum dissolved in \(100 \mathrm{\, g}\) of water. Given the solubility at higher temperatures usually increases, at \(80^{\circ} \mathrm{C}\), this amount \(130 \mathrm{\, g}\) can be assumed to be soluble, resulting in a solution which is likely saturated or unsaturated at this higher temperature since we are not given the solubility at \(80^{\circ} \mathrm{C}\).

Step 3: Analyze the cooled solution

When the solution is cooled to \(50^{\circ} \mathrm{C}\), the solution should, according to the initial solubility at this temperature, only hold \(44 \mathrm{\, g}\) of alum per \(100 \mathrm{\, g}\) of water if it were saturated. However, since it holds \(130 \mathrm{\, g}\), it is supersaturated at \(50^{\circ} \mathrm{C}\), as no precipitate forms initially, which is indicative of such a solution.

Step 4: Induce crystallization

A disturbance, such as scratching with a metal spatula, provides nucleation sites and can disrupt the metastable supersaturated state, causing alum to precipitate out as crystals when the solution is at \(50^{\circ} \mathrm{C}\). This process is called crystallization.

Step 5: Calculate mass of crystals formed

At equilibrium, only \(44 \mathrm{\, g}\) of alum will remain in solution per \(100 \mathrm{\, g}\) of water at \(50^{\circ} \mathrm{C}\), as per its solubility. The excess, given that there were \(130 \mathrm{\, g}\) initially, will precipitate out. Therefore, the mass of precipitated crystals is \(130 \mathrm{\, g} - 44 \mathrm{\, g} = 86 \mathrm{\, g}\).

Key Concepts

Supersaturated Solution

A supersaturated solution occurs when a solution contains more solute than it would normally dissolve at a given temperature. In the case of alum, when you have 130 g dissolved in 100 g of water at 80°C and then cool it to 50°C, it becomes supersaturated at this lower temperature because the solubility limit at 50°C is only 44 g per 100 g of water.

• At 80°C, the additional temperature allows more alum to dissolve.
• As you cool the solution, this excess amount remains dissolved temporarily.
• The solution is unstable and just needs a disturbance to form crystals.

This temporary stability is what makes supersaturated solutions fascinating, as they hold more solute than a simple saturated solution at the same temperature would. This is a crucial aspect in many industrial processes where control of crystallization is important.

Crystallization

Crystallization is the process where a solute comes out of solution to form a solid. In a supersaturated solution, as seen in the cooling of alum, crystallization can be easily initiated with a disturbance, such as scratching the container with a spatula.

• This disturbance acts as a nucleation point or a site where crystal growth starts.
• Crystallization then proceeds as the excess solute finds more sites to form into solid crystals.
• In practical terms, crystallization allows the recovery of solute in solid form from a solution.

The process is widely utilized in various fields such as drug manufacturing, where precise crystallization is necessary to ensure consistent and effective final products.

Equilibrium

Equilibrium in a solution context refers to the state where the rate of solute entering the solution equals the rate of solute leaving it. For alum at 50°C, equilibrium means that it will only hold 44 g in solution per 100 g of water when undisturbed. After crystallization, the solution reaches equilibrium again when the solute concentration returns to this saturated level.

• Initially, the system is not at equilibrium after the cooling because it is supersaturated.
• Once crystallization is initiated, the excess solute precipitates until equilibrium is achieved.
• The resulting equilibrium state will be a saturated solution with the remaining 44 g dissolved.

Understanding equilibrium helps manage solution processes, ensuring the desired amount of solute stays dissolved, while any excess is removed and can be prepared for further use or analysis.