Question

What is the molar chloride ion concentration that results from mixing \(100.0 \mathrm{~mL}\) of \(0.156 \mathrm{M}\) lithium chloride and \(150.0 \mathrm{~mL}\) of \(0.225 \mathrm{M}\) barium chloride?

Short answer

The molar chloride ion concentration is 0.3324 M.

Step-by-step solution

Determine Moles of Chloride from Lithium Chloride

Calculate the moles of chloride ions provided by lithium chloride. Since lithium chloride contributes one chloride ion per molecule, use the formula: moles of chloride = volume (L) \( \times \) molarity. So, \[0.100 \text{ L} \times 0.156 \text{ M} = 0.0156 \text{ moles of chloride ions.}\]

Determine Moles of Chloride from Barium Chloride

Calculate the moles of chloride ions from barium chloride. Barium chloride contributes two chloride ions per molecule. Therefore, \(0.150 \text{ L} \times 0.225 \text{ M} = 0.03375 \text{ moles of barium chloride}\), which contributes \(2 \times 0.03375 = 0.0675\) moles of chloride ions.

Calculate Total Moles of Chloride Ions

Add the moles of chloride ions from both sources: \[0.0156 + 0.0675 = 0.0831 \text{ moles of chloride ions.}\]

Determine Total Volume of the Solution

The total volume of the mixed solution is the sum of both solutions: \[100.0 \text{ mL} + 150.0 \text{ mL} = 250.0 \text{ mL} = 0.250 \text{ L}.\]

Calculate the Molarity of Chloride Ions

Use the number of moles and the total volume to find the molarity: \[\text{Molarity} = \frac{\text{Total moles of chloride ions}}{\text{Total volume in liters}} = \frac{0.0831}{0.250} = 0.3324 \text{ M}.\]

Key concepts

Moles

Understanding moles is fundamental in chemistry, as it allows us to count particles at the atomic or molecular level by using a much larger quantity. A "mole" is just a name for a specific number of particles, similar to how a "dozen" is twelve items, but instead, a mole is precisely Avogadro's number, \[6.022 \times 10^{23}\] particles. The concept of moles helps translate microscopic chemical interactions into amounts we can visualize and measure.

To calculate moles in a solution, you use the relationship between molarity, moles, and volume. The formula is:

• \[\text{Moles} = \text{Molarity} \times \text{Volume (L)}\]

By using this formula, you can determine how many moles of a particular item are present in a solution, which is essential for balancing chemical reactions and calculating concentrations.

Chloride Ion Concentration

When calculating the chloride ion concentration in a solution, it's important to consider the source of these ions. Different compounds, like lithium chloride (\(\text{LiCl}\)) and barium chloride (\(\text{BaCl}_2\)), contribute varying numbers of chloride ions.

For lithium chloride, each molecule dissociates into one lithium ion and one chloride ion in solution:

• 1 LiCl \(\rightarrow\) 1 Li\(^+\) + 1 Cl\(^-\)

Barium chloride, on the other hand, contributes two chloride ions because it dissociates as follows:

• 1 BaCl\(_2\) \(\rightarrow\) 1 Ba\(^{2+}\) + 2 Cl\(^-\)

Understanding this dissociation is critical for accurately calculating the total moles of chloride ions in a solution. By multiplying the moles of the compound by the number of ions produced, you can determine the exact contribution to the chloride ion concentration from each source.

Solution Volume

The volume of a solution is a key factor in determining its molarity, which is the concentration of a solute in a solution. To calculate molarity, you need to know both the total number of moles of solute and the total solution volume in liters.

In practice, solution volumes are often measured in milliliters (mL), but they need to be converted to liters (L) for molarity calculations, as 1 L = 1000 mL. This conversion is straightforward:

• \[\text{Total Volume} = \text{Volume of A} + \text{Volume of B}\]
• \[\text{Total Volume (L)} = \text{Total Volume (mL)} \div 1000\]

For a mixture, the total volume is simply the sum of the volumes of the individual components. Once you know the total volume, you can use it to calculate the final molarity of your ions or solutes in the solution. This approach ensures that you have an accurate understanding of how concentrated a solution is, which is critical for many chemical applications.