Question

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: (a) \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\) or \(0.15 \mathrm{M} \mathrm{KCl}\) solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\) solution or \(400 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{LiCl}\) solution, (c) \(0.050 \mathrm{M} \mathrm{HCl}\) solution or \(0.020 \mathrm{M} \mathrm{CdCl}_{2}\) solution.

Short answer

Higher concentration of Cl⁻ ions for each pair: (a) \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\), (b) \(400\,\mathrm{mL}\) of \(0.080\, \mathrm{M} \mathrm{LiCl}\) solution, and (c) \(0.050\, \mathrm{M} \mathrm{HCl}\) solution.

Step-by-step solution

Pair (a) - Molarities

(a) \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\) or \(0.15 \mathrm{M} \mathrm{KCl}\) For \(\mathrm{CaCl}_{2}\), there are 2 Cl⁻ ions in each molecule. For \(\mathrm{KCl}\), there is 1 Cl⁻ ion in each molecule.

Pair (a) - Calculating concentration of Cl⁻ ions

For \(\mathrm{CaCl}_{2}\): \(0.10 \mathrm{M}\times2 = 0.20 \mathrm{M}\) Cl⁻ ion. For \(\mathrm{KCl}\): \(0.15 \mathrm{M}\times1= 0.15 \mathrm{M}\) Cl⁻ ion. Higher concentration of Cl⁻ ions: \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\)

Pair (b) - Molarities

(b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\) solution or \(400 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{LiCl}\) solution. For \(\mathrm{KCl}\), there is 1 Cl⁻ ion in each molecule. For \(\mathrm{LiCl}\), there is 1 Cl⁻ ion in each molecule.

Pair (b) - Calculating concentration of Cl⁻ ions in the given volumes

For \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\): \(0.10 \mathrm{M}\times1\times100 \mathrm{~mL}= 10\,\mathrm{mmol}\) Cl⁻ ion. For \(400 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{LiCl}\): \(0.080 \mathrm{M}\times1\times400\mathrm{~mL} = 32\,\mathrm{mmol}\) Cl⁻ ion. Higher concentration of Cl⁻ ions: \(400\,\mathrm{mL}\) of \(0.080\, \mathrm{M} \mathrm{LiCl}\) solution.

Pair (c) - Molarities

(c) \(0.050\,\mathrm{M} \mathrm{HCl}\) solution or \(0.020\,\mathrm{M} \mathrm{CdCl}_{2}\) solution. For \(\mathrm{HCl}\), there is 1 Cl⁻ ion in each molecule. For \(\mathrm{CdCl}_{2}\), there are 2 Cl⁻ ions in each molecule.

Pair (c) - Calculating concentration of Cl⁻ ions

For \(\mathrm{HCl}\): \(0.050\,\mathrm{M}\times1 = 0.050\,\mathrm{M}\) Cl⁻ ion. For \(\mathrm{CdCl}_{2}\): \(0.020\,\mathrm{M}\times2 = 0.040\,\mathrm{M}\) Cl⁻ ion. Higher concentration of Cl⁻ ions: \(0.050\, \mathrm{M} \mathrm{HCl}\) solution.

Key concepts

Molarity

Molarity is an essential concept in chemistry, particularly when dealing with solutions. It is a way to express the concentration of a solution in terms of the amount of substance per volume of solution. Specifically, molarity (symbolized as M) is defined as the number of moles of solute present in one liter of solution. This can be written mathematically as:\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\]Understanding molarity is crucial for accurately preparing solutions and performing concentration calculations. It allows chemists to predict how solutions will react in various conditions. Moreover, molarity directly influences the reaction rate and equilibrium in chemical processes. For students, grasping this concept means being able to communicate about substances in a language that is standardized across the scientific community.
For example, in the exercise above, when comparing solutions like \(0.10\ \mathrm{M}\ \mathrm{CaCl}_{2}\) and \(0.15\ \mathrm{M}\ \mathrm{KCl}\), knowing the molarity helps determine the amount of chloride ions contributed by each compound, as each formula unit provides different ions.

Chemical solutions

Chemical solutions consist of solutes dissolved in solvents, creating homogeneous mixtures at the molecular level. In these solutions, the solute is the substance that gets dissolved, while the solvent is the material that dissolves the solute.
Chemical solutions are foundational to various processes in chemistry, ranging from simple dissolution to complex reaction mixtures. The composition of a solution, notably including its molarity, directly affects its properties, such as boiling and freezing points, vapor pressure, and osmotic pressure.

• Solute: The substance dissolved, like \(\mathrm{CaCl}_{2}\), \(\mathrm{KCl}\), or \(\mathrm{HCl}\) in the examples.
• Solvent: Typically, the substance in greater amount, often water in laboratory solutions.
In a laboratory setting, chemical solutions can be manipulated by adjusting concentrations to study effects on chemical reactions or to prepare reagents precisely for experimental needs. For instance, by understanding the role of each component, we can predict the behavior of solutions like those in the exercise, determining which has higher solute concentrations and thus more intense chemical activities.

Concentration calculations

Concentration calculations are a vital aspect of chemistry that involve determining how much solute is present in a given volume of solution. These calculations allow chemists to modify and understand the properties of solutions, making them integral to experimental designs and industrial applications.
The formulas used in concentration calculations depend on the type of concentration measurement needed. For molarity, as highlighted earlier, we calculate based on moles per liter. But in particular scenarios, such as the exercise, we also consider the stoichiometry of the dissolved solutes:

• For ionic compounds like \(\mathrm{CaCl}_{2}\), the formula unit's stoichiometric coefficient tells us that it releases two \(\mathrm{Cl}^{-}\) ions per formula unit.
• Conversely, \(\mathrm{KCl}\) releases only one \(\mathrm{Cl}^{-}\) ion per formula unit.

By multiplying the molarity by the number of ions produced in solution (as with \(\mathrm{CdCl}_{2}\) releasing two \(\mathrm{Cl}^{-}\) ions), we accurately determine the concentration of specific ions. Practicing such calculations enhances one's ability to makes predictions and adjustments in both experimental and theoretical contexts, making it a cornerstone of chemical analysis and synthesis.