Problem 26 - Chemistry: The Central Science

Question

By referring to Figure 13.15, determine the mass of each of the following salts required to form a saturated solution in \(250 \mathrm{~g}\) of water at \(30^{\circ} \mathrm{C}\) : (a) \(\mathrm{KClO}_{3}\), (b) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Ce}_{2}\left(\mathrm{SO}_{4}\right)_{3}\).

Step-by-Step Solution

Answer

To calculate the mass of each salt needed to form a saturated solution in 250g of water at 30°C, use the solubility values from Figure 13.15 (not provided here) and set up a proportion: \( mass\_of\_salt= solubility\_value \times \dfrac{mass\_of\_water}{100} \) Replace solubility_value with the values from Figure 13.15 and mass_of_water with 250g for each salt (KClO3, Pb(NO3)2, and Ce2(SO4)3) and perform the calculations.

Step 1: Find the solubility of each salt from Figure 13.15 at 30°C

Given the graph in Figure 13.15, find the solubility values at 30°C for KClO3, Pb(NO3)2, and Ce2(SO4)3. The solubility values will be in g/100g of water.

Step 2: Calculate the mass of each salt for 250g of water

To determine the mass of each salt required for 250g of water, set up a proportion: \( mass\_of\_salt= solubility\_value \times \dfrac{mass\_of\_water}{100} \) Where mass_of_salt is the mass of the respective salt in grams required for a saturated solution, solubility_value is the solubility value from Step 1, and mass_of_water is given as 250g. (a) For KClO3: \( mass\_of\_KClO3= solubility\_value \times \dfrac{250}{100} \) (b) For Pb(NO3)2: \( mass\_of\_Pb(NO3)2= solubility\_value \times \dfrac{250}{100} \) (c) For Ce2(SO4)3: \( mass\_of\_Ce2(SO4)3= solubility\_value \times \dfrac{250}{100} \)

Step 3: Perform the calculations

Substitute the solubility values from Figure 13.15 (Step 1) into the equations from Step 2 and calculate the mass of each salt required for the saturated solution in 250g of water at 30°C. Note: Since the Figure 13.15 is not provided here, the actual solubility values and mass calculations cannot be performed. However, following the above steps with the solubility values would yield the correct mass of each salt for a saturated solution in 250g of water at 30°C.

Key Concepts

Saturated Solution Chemistry

Understanding saturated solution chemistry is fundamental in the study of solutes and solvents. A saturated solution is one which holds the maximum amount of solute that can be dissolved by the solvent at a given temperature and pressure. Any additional solute will not dissolve in the solution and will remain in a separate phase, either as a precipitate at the bottom of a container or as crystals within the solution.

Temperature plays a crucial role in solubility; generally, as temperature increases, the solubility of solids in a liquid also increases. However, this is not a strict rule for all solutes. The formation of a saturated solution can be observed in everyday life, such as when sugar or salt is added to water until no more can dissolve, resulting in undissolved material settling at the bottom.

In the context of an educational exercise, understanding the chemistry behind saturated solutions allows students to make precise measurements, predictions, and adjustments based on solubility data. It's not simply about getting the right answer; it's about grasping the interplay between temperature, solubility, and concentration.

Mass Proportion Calculations

Mass proportion calculations are integral to preparing specific concentrations of solutions, including saturated solutions. To carry out these calculations, one often sets up a proportion based on the known solubility of a substance. The formula to calculate the mass of the solute needed to create a saturated solution in a particular amount of solvent is given by:

\[\begin{equation} mass\_of\_solute = solubility\_value \times \dfrac{mass\_of\_solvent}{100} \end{equation}\]
This is based on the solubility being expressed in terms of grams of solute per 100 grams of solvent, which is a common convention in chemistry. When the mass of the solvent deviates from 100 grams, a direct proportion is used to find the corresponding mass of the solute that will create a saturated solution.

Solving practical examples enhances the student's ability to manipulate proportions and understand the quantitative relationship between solute and solvent. It's a vital skill not only in chemistry but across various scientific and engineering disciplines.

Solubility Curves Interpretation

Interpreting solubility curves is essential for accurately determining the solubility of a substance at a specific temperature. A solubility curve is a graph that shows the relationship between solubility (y-axis) and temperature (x-axis) for a given substance in a particular solvent. These curves are useful for predicting how much solute can be dissolved in a solvent at any temperature within a range.

For instance, when given a solubility curve for different salts, like those in the exercise, students can identify the exact solubility at 30°C by finding the corresponding point on the curve. If you move horizontally from the 30°C marker until you reach the curve for the salt in question, then move vertically down, the value where you meet the x-axis tells you the solubility in grams per 100 grams of solvent.

These graphical tools are invaluable for not only solving homework problems but also in real-world applications where predicting concentrations at various temperatures can affect the outcome of chemical processes. For students, mastering the interpretation of these curves is an empowering analytical tool in chemistry.