Determine the concentration of \(\mathrm{Cl}^{-}\)in each aqueous solution. (Assume complete dissociation of each compound.) (a) \(0.15 \mathrm{M} \mathrm{NaCl}\) (b) \(0.15 \mathrm{MCuCl}_{2}\) (c) \(0.15 \mathrm{M} \mathrm{AlCl}_{3}\)

When we talk about the concentration of an aqueous solution, we refer to the amount of a substance dissolved in a certain volume of water. This is typically measured in molarity (M), which is moles of solute per liter of solution. Understanding this concept is fundamental for determining how much of a given ion is present in a solution, especially after dissolution.

In the context of the exercise, the given concentrations are all initially at 0.15 M, meaning there are 0.15 moles of the compound (whether it is NaCl, CuCl₂, or AlCl₃) in one liter of water. Upon dissolution, these compounds release ions into solution, which changes the concentration of the respective ions according to the compound's stoichiometry.

Stoichiometry is the part of chemistry that studies the quantitative relationships or ratios between reactants and products in a chemical reaction. It is also applied when determining the relationship between different ions in a solution after a compound dissociates.

For instance, the stoichiometry of NaCl in the exercise indicates that for each mole of NaCl that dissolves, one mole of Na⁺ ions and one mole of Cl^- ions is formed. Similarly, stoichiometry for CuCl₂ and AlCl₃ shows that as these compounds dissolve, they release two and three times as many chloride ions per formula unit, respectively. This information allows us to calculate the resulting concentrations of chloride ions in solution after dissociation.

Ionic compounds, such as NaCl, CuCl₂, and AlCl₃ mentioned in the exercise, are composed of positively and negatively charged ions. These compounds are held together by the strong electrostatic forces between oppositely charged ions. In water, ionic compounds tend to dissociate into their constituent ions.

Under the assumption of complete dissociation, which is common for many salts in a dilute aqueous solution, each ionic compound will separate entirely into cations and anions. NaCl dissociates into Na⁺ and Cl^-, CuCl₂ into Cu²⁺ and twice as many Cl^- due the subscript '2', and AlCl₃ splits into Al³⁺ and three times as many Cl^- ions. Recognizing these patterns is essential not just for understanding how compounds behave in solution, but also for predicting the conductivity, osmotic pressure, and reactivity of the solution.