Consider the insoluble salts \(\mathrm{JQ} \mathrm{K}_{2} \mathrm{R}, \mathrm{L}_{2} \mathrm{~S}_{3}, \mathrm{MT}_{2}\), and \(\mathrm{NU}_{3}\). They are formed from the metal ions \(\mathrm{J}^{+}, \mathrm{K}^{+}, \mathrm{L}^{3+}, \mathrm{M}^{2+}\), and \(\mathrm{N}^{3+}\) and the nonmetal ions \(\mathrm{Q}^{-}, \mathrm{R}^{2-}, \mathrm{S}^{2-}, \mathrm{T}^{-}\), and \(\mathrm{U}^{-}\). All the salts have the same \(K_{\text {sp }}, 1 \times 10^{-10}\), at \(25^{\circ} \mathrm{C}\). (a) Which salt has the highest molar solubility? (b) Does the salt with the highest molar solubility have the highest solubility in g salt/100 g water? (c) Can the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water be determined from the information given? If yes, calculate the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water. If no, why not?

The solubility product constant, or Ksp, is fundamental in understanding the solubility of sparingly soluble salts. It represents the level at which a salt can dissolve in a solution before reaching a point of equilibrium, where the rate at which the salt dissolves equals the rate at which it precipitates. The numerical value of Ksp is specific to each salt and gives us insight into how soluble a substance is; the lower the Ksp, the less soluble the salt is.

For salts yielding multiple ions upon dissolution, Ksp can be represented using the formula: \[K_{sp} = [Cation]^{m}[Anion]^{n}\] where m and n are the stoichiometric coefficients of the cation and anion in the balanced dissolution equation. For example, if a salt AB2 dissolves to form A^2+ and 2B^-, the Ksp expression would be \[K_{sp} = [A^{2+}][B^-]^{2}\].

When given Ksp and the dissolution equation, we can calculate the molar solubility of the salt, which is the maximum amount of the salt that can dissolve in a solution to form a saturated solution. The exercises and solutions provided play with this concept to determine which of the given salts would have the highest molarity in a saturated solution based on their common Ksp.

Chemical equilibrium plays a significant role when discussing solubility. At equilibrium, the rate of dissolution (substance dissolving into ions) equals the rate of precipitation (ions coming together to form the substance). This does not mean that the amounts of dissolved and undissolved substances are equal, but that their rates of interconversion are, leading to a stable concentration of ions in solution.

When we dissolve an insoluble salt, it starts dissociating into ions until it reaches a point where the dissolved ions can saturate the solution. Beyond this saturation point, any additional salt will not dissolve and will remain as a solid, maintaining a dynamic equilibrium between the solid and dissolved ions. By understanding and applying the principles of chemical equilibrium, we can predict whether adding more salt will dissolve or if it will simply add to the undissolved amount.

Solubility conversion is an essential step when comparing the solubility of different salts, especially in real-world applications like chemistry experiments and pharmaceutical formulations. Molar solubility, which is often calculated from the Ksp value, tells us how many moles of a salt can dissolve in a liter of water to form a saturated solution. However, this information is not always practical because laboratories and industries commonly use weight measurements.

To convert molar solubility to solubility in grams per 100 g of water (mass solubility), we multiply the molar solubility by the molar mass of the salt. For example, if the molar solubility of a salt XY is \[x \text{ mol/L}\], and the molar mass of XY is M g/mol, then the solubility in g/100g water is given by: \[\text{Solubility (g/100g water)} = x \times M \times \frac{100}{1000}\].

Through this conversion, we can compare the solubility of different salts on a mass basis, which is much more intuitive for understanding how much of the substance can actually be dissolved in a certain amount of water. This is particularly useful when compounds have very different molar masses, as it provides a common ground for comparison.