Question

The solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) at \(30^{\circ} \mathrm{C}\) is \(0.15 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water. Describe the solution formed when \(1.0 \mathrm{~g} \mathrm{Ca}(\mathrm{OH})_{2}\) is added to \(100.0 \mathrm{~g}\) of water.

Short answer

The solution will be a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water, with \(0.85 \mathrm{~g}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) forming a precipitate because the solubility is exceeded.

Step-by-step solution

Analyze the provided solubility value

The solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) given is \(0.15 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water. This indicates the maximum mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that can be dissolved in \(100 \mathrm{~g}\) of water without forming any precipitate.

Compare the added mass with the solubility

Compare the mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that has been added, which is \(1.0 \mathrm{~g}\), with its maximum solubility in \(100.0\mathrm{~g}\) water, which is \(0.15 \mathrm{~g}\). Since the added mass is greater than the maximum solubility, it indicates that the solution will be a saturated one with some precipitate of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Calculate the precipitate formed

The mass of the excess substance which cannot dissolve in water is the mass of the precipitated \(\mathrm{Ca}(\mathrm{OH})_{2}\). That can be calculated as \(1.0 \mathrm{~g} - 0.15 \mathrm{~g} = 0.85 \mathrm{~g}\). Thus, \(0.85 \mathrm{~g}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) will precipitate out of the solution.

Key concepts

Saturated Solution

Understanding the concept of a saturated solution is essential in chemistry, particularly when exploring solubility and reactions in solution. A saturated solution refers to a solution that has reached the maximum concentration of solute that it can dissolve at a given temperature and pressure. At this point, any additional solute will remain undissolved and may precipitate, or settle out as a solid.

Solubility, the measure of how much solute can be dissolved in a solvent, varies depending on the substances involved and the conditions of the solution. In our example, \(\mathrm{Ca}(\mathrm{OH})_{2}\) has a solubility of \(0.15 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water at \(30^\circ \mathrm{C}\). When \(1.0 \mathrm{~g}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is added to \(100.0 \mathrm{~g}\) of water, this exceeds its solubility, leading to a saturated solution with excess \(\mathrm{Ca}(\mathrm{OH})_{2}\) forming a precipitate.

It's key for students to visualize this concept. Imagine adding sugar to a glass of water and stirring it. At first, the sugar dissolves, but as you add more, it begins to accumulate at the bottom of the glass. That’s because the water can only dissolve a certain amount of sugar, beyond which it becomes saturated and the sugar settles as a precipitate.

Precipitation Reactions

Precipitation reactions are central to understanding how chemical species interact in aqueous solutions. These reactions occur when ions in solution combine to form an insoluble compound that falls out of solution as a solid, known as a 'precipitate'.

To identify a precipitation reaction, one must know the solubility rules, which predict the solubility of different ionic compounds. In our textbook example, when more \(\mathrm{Ca}(\mathrm{OH})_{2}\) is added than can be dissolved, it causes a precipitation reaction. The \(\mathrm{Ca}(\mathrm{OH})_{2}\) that cannot be held in solution forms a solid, settling out of the solution.

Recognizing these reactions is pivotal for students when conducting experiments or solving problems involving ionic compounds. To reinforce the concept, one might consider conducting a hands-on experiment where a solution is made supersaturated on purpose by slowly adding solute, thus visually demonstrating the creation of a precipitate when the substance's solubility limit is exceeded.

Stoichiometry

Another fundamental concept in chemistry is stoichiometry, which deals with the quantitative relationship between the reactants and products in a chemical reaction. Stoichiometry allows chemists to predict the amounts of substances consumed and produced in a given reaction based on the balanced chemical equation.

In the context of our problem, where \(\mathrm{Ca}(\mathrm{OH})_{2}\) is being dissolved in water, stoichiometry is not directly applied as it is a simple dissolution rather than a chemical reaction. However, when considering the formation of the precipitate, one can think of stoichiometry in terms of the mass balance - comparing the mass of solute added to the mass that can actually dissolve, and thus determining the mass of the precipitate. In our step by step solution, the calculation \(1.0 \mathrm{~g} - 0.15 \mathrm{~g} = 0.85 \mathrm{~g}\) follows the stoichiometric principles of mass conservation.

For educational purposes, students should practice stoichiometry through a variety of problems, such as predicting yields of chemical reactions and converting between masses, moles, and molecules, to become proficient in this area of chemistry.