After \(50.0 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) solution of calcium nitrate is combined with \(100.0 \mathrm{~mL}\) of a \(0.835 \mathrm{M}\) solution of calcium nitrate, \((\) a) what is the molar concentration of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) in the combined solution? Once in solution, the calcium nitrate exists not as intact calcium nitrate but rather as calcium ions and nitrate ions. What are the molar concentrations (b) of \(\mathrm{Ca}^{2+}(a q)\) in the combined solution and (c) of \(\mathrm{NO}_{3}^{-}(a q)\) in the combined solution?

Solution concentration is a measure of the amount of solute present in a given quantity of solvent or solution. It's critical for students to grasp this concept as it's the basis for understanding chemical reactions that take place in solutions. One of the most common units for expressing concentration is molarity, which is signified by the symbol 'M' and defined as moles of solute per liter of solution.

To calculate the molarity, you need to know the number of moles of solute and the volume of the solution in liters. The well-known formula is depicted as:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
This knowledge is vital when performing dilutions or combining solutions, as seen in the exercise. Understanding how to combine the molarities of two solutions, while accounting for changes in volume, allows for accurate prediction of final concentrations in the chemical contexts such as reactions and biological assays.

Chemical molarity is a specific term for the concentration of a solution, where the solute is a chemical substance. It's a measure of how much of a chemical is present in a given volume of solution, which is especially important in chemistry because it reflects the 'strength' or 'potency' of a solution to react or participate in chemical processes.

For students tackling molarity calculations, it's crucial to work with volumetric measurements (like milliliters and liters) and the concept of moles. An example of a molarity calculation, based on the exercise provided, involves the use of the formula: \[ M_1V_1 + M_2V_2 = M_fV_f \]
This equation helps to find the final molarity after mixing two solutions with different concentrations and volumes. For instance, if two solutions of calcium nitrate with known molarities are mixed, their volumes add up, and their molarities must be appropriately weighed to find the new molarity—keeping in mind that volume should always be considered in liters to maintain consistency in molarity units.

When ionic compounds dissolve in water, they dissociate into their respective ions, a fact that is significant for understanding the behavior of solutions at the ionic level. In the given exercise, calcium nitrate dissociates into calcium ions (\( \mathrm{Ca}^{2+} \)) and nitrate ions (\( \mathrm{NO}_{3}^{-} \)).

It's key to note that the molarity of the ions in solution might differ from the molarity of the compound due to the stoichiometry of the dissociation process. For instance:

• Each formula unit of \( \mathrm{Ca(NO_{3})_{2}} \) produces one \( \mathrm{Ca}^{2+} \) ion in solution.
• Each formula unit also yields two \( \mathrm{NO_{3}}^{-} \) ions.

The molarity of the ions is then directly related to the molarity of the dissolved compound and the ion-to-formula unit ratio. In the exercise solution, the final molarity of calcium ions is equal to the molarity of the calcium nitrate, whereas the molarity of nitrate ions is double because two nitrate ions are produced for every formula unit of calcium nitrate.