The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per \(100 \mathrm{mL}\) of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) \(\mathrm{BaSiF}_{6}, 0.026 \mathrm{g} / 100 \mathrm{mL}\) (contains \(\mathrm{SiF}_{6}^{2-}\) ions) (b) \(\operatorname{Ce}\left(\mathrm{IO}_{3}\right)_{4}, 1.5 \times 10^{-2} \mathrm{g} / 100 \mathrm{mL}\) (c) \(\mathrm{Gd}_{2}\left(\mathrm{SO}_{4}\right)_{3}, 3.98 \mathrm{g} / 100 \mathrm{mL}\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtBr}_{6}, 0.59 \mathrm{g} / 100 \mathrm{mL}\) (contains \(\mathrm{PtBr}_{6}^{2-}\) ions)

Solubility refers to the maximum amount of a substance, known as the solute, that can dissolve in a specific amount of solvent at a given temperature and pressure to form a homogeneous mixture called a solution. In the context of your exercise, solubility is expressed in grams of solute per 100 mL of water.

In chemistry, solutes can be gases, liquids, or solids, and solvents are most commonly liquids. Water is known as the 'universal solvent' because of its ability to dissolve many substances. However, not all solutes dissolve equally in water due to factors such as ionic strength, temperature, and the presence of other chemicals that could change the solubility of the solute.

To understand how a slightly soluble compound behaves, we assume that the change in volume upon dissolution is negligible. This assumption simplifies the calculation of the solubility product, a special equilibrium constant denoted as K_sp. This constant helps predict the solubility of a compound under various conditions.

A dissolution equation represents the process of a solute dissolving in a solvent to form ions in a solution. It is a critical aspect of solubility calculations. When a compound dissolves in water, it may dissociate into its respective ions.

For instance, the dissolution of barium hexafluorosilicate, BaSiF₆, is represented by the equation:\[BaSiF_{6}(s) \rightleftharpoons Ba^{2+}(aq) + SiF_{6}^{2-}(aq)\].In this equilibrium expression, the solid (s) on the left of the arrow represents the undissolved solute, and the aqueous (aq) ions on the right are the products of the solute dissolving in water.

The arrow pointing in both directions signifies that the dissolution process is reversible and has reached a state of dynamic equilibrium. At this point, the rate of dissolution is equal to the rate of precipitation, which is critical for determining the solubility product constant, K_sp.

The molar mass is the weight of one mole (Avogadro's number, or 6.022 x 10²³ particles) of a substance, normally expressed in grams per mole (g/mol). It is fundamental in converting between mass and mole, which allows chemists to calculate the concentration of substances in solution.

To calculate the molar mass of a compound, you add up the atomic masses of each element, as found on the periodic table, multiplied by the number of atoms of each element in the compound's formula. For example, Barium Hexafluorosilicate (BaSiF₆) would have a molar mass calculated as follows:\[ M_{BaSiF_{6}} = (1 \times M_{Ba}) + (1 \times M_{Si}) + (6 \times M_{F}) \],where M represents the atomic mass of each element. This step is pivotal because you need the molar mass to determine the molarity of the solution.

Molarity is a measure of the concentration of a solute in a solution, defined as the moles of solute per liter of solution (mol/L). It is a crucial concept for scientists when making solutions in a lab because it allows for precise calculations of reactants and products in chemical reactions.

To calculate molarity, you divide the amount of solute in moles by the volume of solution in liters:\[ Molarity (M) = \frac{moles \text{ of solute}}{liters \text{ of solution}} \].In your exercise, to convert grams to moles, you use the molar mass of the solute. It is important to remember that the solubility provided in the exercise is based on 100 mL of water, so to find the molarity, you will also need to convert this volume to liters. This concentration, expressed in molarity, then feeds directly into the calculation of the solubility product for the compound in question.