Problem 70 - Chemistry: The Central Science

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, \((\mathrm{c}) 0.050 \mathrm{M} \mathrm{HCl}\) solution or \(0.020 \mathrm{M} \mathrm{CdCl}_{2}\) solution.

Step-by-Step Solution

Answer

(a) The 0.10 M \(\mathrm{CaCl}_{2}\) solution has a higher concentration of \(\mathrm{Cl}^-\) ions (0.20 M) compared to the 0.15 M \(\mathrm{KCl}\) solution (0.15 M). (b) The 100 mL of 0.10 M \(\mathrm{KCl}\) solution has a higher concentration of \(\mathrm{Cl}^-\) ions (0.10 M) compared to the 400 mL of 0.080 M \(\mathrm{LiCl}\) solution (0.080 M). (c) The 0.050 M \(\mathrm{HCl}\) solution has a higher concentration of \(\mathrm{Cl}^-\) ions (0.050 M) compared to the 0.020 M \(\mathrm{CdCl}_{2}\) solution (0.040 M).

Step 1: (a) Compare 0.10 M \(\mathrm{CaCl}_{2}\) and 0.15 M \(\mathrm{KCl}\) solutions

: For every 1 mole of \(\mathrm{CaCl}_{2}\), there are 2 moles of \(\mathrm{Cl}^-\). So, in a 0.10 M \(\mathrm{CaCl}_{2}\) solution, the concentration of \(\mathrm{Cl}^-\) ions is: \(0.10\ \mathrm{M}\ \mathrm{CaCl}_{2} \times 2 = 0.20\ \mathrm{M}\) The concentration of \(\mathrm{Cl}^-\) ions in the 0.15 M \(\mathrm{KCl}\) solution is equal to the concentration of the \(\mathrm{KCl}\) since there is only 1 mole of \(\mathrm{Cl}^-\) for every mole of \(\mathrm{KCl}\). So, the concentration of \(\mathrm{Cl}^-\) ions is: \(0.15\ \mathrm{M}\) Comparing the concentrations: \(0.20\ \mathrm{M} > 0.15\ \mathrm{M}\) Thus, the \(\textbf{0.10 M}\ \mathrm{CaCl}_{2}\ \textbf{solution}\) has the higher concentration of \(\mathrm{Cl}^-\) ions.

Step 2: (b) Compare 100 mL of 0.10 M \(\mathrm{KCl}\) and 400 mL of 0.080 M \(\mathrm{LiCl}\) solutions

: First, find the moles of \(\mathrm{Cl}^-\) ions in each solution: 100 mL of 0.10 M \(\mathrm{KCl}\) solution: \(\mathrm{moles} \ \mathrm{Cl}^- = \mathrm{volume} \times \mathrm{concentration}\) \(\mathrm{moles} \ \mathrm{Cl}^- = 100\ \mathrm{mL} \times 0.10\ \mathrm{M} = 10\ \mathrm{mmol}\) 400 mL of 0.080 M \(\mathrm{LiCl}\) solution: \(\mathrm{moles} \ \mathrm{Cl}^- = \mathrm{volume} \times \mathrm{concentration}\) \(\mathrm{moles} \ \mathrm{Cl}^- = 400\ \mathrm{mL} \times 0.080\ \mathrm{M} = 32\ \mathrm{mmol}\) Now, compare the mole amounts and determine the concentration in each solution: \(10/100\ \mathrm{mL} = 0.10\ \mathrm{M}\) \(32/400\ \mathrm{mL} = 0.080\ \mathrm{M}\) Comparing the concentrations: \(0.10\ \mathrm{M} > 0.080\ \mathrm{M}\) Thus, the \(\textbf{100 mL of 0.10 M}\ \mathrm{KCl}\ \textbf{solution}\) has the higher concentration of \(\mathrm{Cl}^-\) ions.

Step 3: (c) Compare 0.050 M \(\mathrm{HCl}\) and 0.020 M \(\mathrm{CdCl}_{2}\) solutions

: The concentration of \(\mathrm{Cl}^-\) ions in the 0.050 M \(\mathrm{HCl}\) solution is equal to the concentration of the \(\mathrm{HCl}\) since there is only 1 mole of \(\mathrm{Cl}^-\) for every mole of \(\mathrm{HCl}\). So, the concentration of \(\mathrm{Cl}^-\) ions is: \(0.050\ \mathrm{M}\) For every 1 mole of \(\mathrm{CdCl}_{2}\), there are 2 moles of \(\mathrm{Cl}^-\) ions. So, in a 0.020 M \(\mathrm{CdCl}_{2}\) solution, the concentration of \(\mathrm{Cl}^-\) ions is: \(0.020\ \mathrm{M}\ \mathrm{CdCl}_{2} \times 2 = 0.040\ \mathrm{M}\) Comparing the concentrations: \(0.050\ \mathrm{M} > 0.040\ \mathrm{M}\) Thus, the \(\textbf{0.050 M}\ \mathrm{HCl}\ \textbf{solution}\) has the higher concentration of \(\mathrm{Cl}^-\) ions.

Key Concepts

Molarity

Molarity is a key concept in chemistry that refers to the concentration of a solution, expressing how many moles of a solute are present in one liter of solution. It is represented by the formula \( M = \frac{n}{V} \), where \( M \) is the molarity, \( n \) is the number of moles, and \( V \) is the volume of the solution in liters. Understanding molarity is crucial when comparing the concentrations of ions in different chemical solutions.To determine the concentration of a particular ion in a solution, it's essential to consider the formula/unit of the compound. For example:

• In \( \mathrm{CaCl}_2 \), each molecule provides two chloride ions (\(\mathrm{Cl}^-\)).
• In \( \mathrm{KCl} \), each molecule provides one chloride ion.
• In \( \mathrm{HCl} \), each molecule provides one chloride ion.
• In \( \mathrm{CdCl}_2 \), each molecule provides two chloride ions.

This insight helps in calculating the molarity of ions like \( \mathrm{Cl}^- \) in different solutions by straightforward multiplication, given the molarity of the compound in question.

Chemical Solutions

Chemical solutions are homogeneous mixtures composed of solutes and solvents. The solute is the substance dissolved, while the solvent is the substance in which the solute dissolves. When preparing solutions, the concentration of ions can be very important, especially when it comes to reactions and properties of the solution.In practice, the type of solute and how it dissociates in the solution matters greatly. For example:

• \( \mathrm{CaCl}_2 \), when dissolved, dissociates into 3 ions (1 \( \mathrm{Ca}^{2+} \) and 2 \( \mathrm{Cl}^- \)).


• \( \mathrm{KCl} \) dissociates into 2 ions (1 \( \mathrm{K}^+ \) and 1 \( \mathrm{Cl}^- \)).

This dissociation affects the concentration of ions. Understanding this principle is crucial when calculating the concentration of specific ions, such as \( \mathrm{Cl}^- \), in any chemical solution.

Concentration Comparisons

Understanding and comparing the concentration of ions within solutions involves more than just examining the molarity. You should also consider how solutes dissociate into ions within the solvent. This is why knowing the dissociation properties of different chemical compounds is important.When presented with two solutions and tasked to determine which one has a higher concentration of \( \mathrm{Cl}^- \) ions, the following steps can be applied:

• Identify the number of \( \mathrm{Cl}^- \) ions released by the solute in the solution. This knowledge is crucial for an accurate comparison.


• Multiply the molarity of the solution by the number of \( \mathrm{Cl}^- \) ions produced per formula unit of the solute. This gives the actual molarity of the \( \mathrm{Cl}^- \) ions in the solution.


• Compare these values across different solutions to determine the higher concentration.

By implementing this process, you can more accurately assess which solution contains a greater concentration of chloride ions, enhancing your understanding of chemical solutions and their behaviors.