Potassium sulfate has a solubility of \(15 \mathrm{~g} / 100 \mathrm{~g}\) water at \(40^{\circ} \mathrm{C}\). A solution is prepared by adding \(39.0 \mathrm{~g}\) of potassium sulfate to \(225 \mathrm{~g}\) of water, carefully heating the solution, and cooling it to \(40^{\circ} \mathrm{C}\). A homogeneous solution is obtained. Is this solution saturated, unsaturated, or supersaturated? The beaker is shaken, and precipitation occurs. How many grams of potassium sulfate would you expect to crystallize out?

Understanding the nature of a supersaturated solution is vital for comprehending certain chemical processes. A supersaturated solution occurs when a solution contains more dissolved solute than it should be able to according to its temperature-based solubility limit.

In the scenario with potassium sulfate, the solution becomes supersaturated when it holds more than 33.75 grams of potassium sulfate per 225 grams of water at 40°C. The solution originally prepared in the exercise indeed has 39 grams of potassium sulfate, exceeding the solubility and thus resulting in a supersaturated solution.

Several factors can lead to the formation of such a solution. One common method is by heating the solvent, which in most cases increases the solubility of a solute. When this hot solution is carefully cooled, the solute doesn't immediately precipitate out. It's a delicate state, where the solution is holding more solute than it 'should,' and it's often just a disturbance away from solute precipitation.

Solute precipitation happens when a solute exceeds its solubility in a solvent and begins to form a solid. For potassium sulfate at 40°C, any amount over 15 grams per 100 grams of water exceeds solubility and is expected to precipitate.

In conditions of a supersaturated solution, like the one mentioned, the addition of further solute or a disturbance such as shaking or scratching the vessel often triggers this precipitation. The solute molecules start to cluster together, forming solid crystals that can no longer be retained in the solution.

In the context of this problem, after the beaker is shaken, it should theoretically result in the precipitation of the excess 5.25 grams of potassium sulfate since that is the amount surpassing the solubility limit at the given temperature.

Solubility calculations provide quantitative insights into how much solute can dissolve in a solvent at a particular temperature. For the potassium sulfate example, these calculations involve the application of ratios and unit conversions.

A direct proportion between the amount of solute able to dissolve in a reference amount of solvent and the amount solvable in the given quantity of solvent is used to determine solubility in different conditions. It's mathematically represented as: \( \frac{{15 \mathrm{{g}}}}{{100 \mathrm{{g}}}} = \frac{{x \mathrm{{g}}}}{{225 \mathrm{{g}}}} \)

Upon solving for x (the unknown amount of the solute soluble in the given solvent mass), we find that only 33.75 grams of potassium sulfate is the maximum solubility for 225 grams of water. Such calculations are essential for predicting whether a solution will be unsaturated, saturated, or supersaturated under specific conditions.