Problem 20 - Analytical Chemistry

Question

What is the molar concentration of \(\mathrm{Cl}^{-}\) in a solution prepared by mixing \(25.0 \mathrm{~mL}\) of \(0.025 \mathrm{M} \mathrm{NaCl}\) with \(35.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{BaCl}_{2}\) ? What is pCl for the mixture?

Step-by-Step Solution

Answer

The molar concentration of Cl⁻ is 0.06875 M, and pCl is approximately 1.16.

Step 1: Calculate moles of Cl⁻ from NaCl

The molar concentration of NaCl is given as 0.025 M, and the volume of the solution is 25.0 mL, which is 0.025 L. Use the formula: \[\text{moles of } \mathrm{Cl}^{-} = (\text{concentration})(\text{volume}) = 0.025 \text{ M} \times 0.025 \text{ L} = 0.000625 \text{ moles}\]

Step 2: Calculate moles of Cl⁻ from BaCl₂

The molar concentration of BaCl₂ is given as 0.050 M, and the volume of the solution is 35.0 mL, which is 0.035 L. BaCl₂ dissociates to give 2 chloride ions per formula unit. Therefore:\[\text{moles of } \mathrm{Cl}^{-} = 2 \times (\text{concentration})(\text{volume}) = 2 \times 0.050 \text{ M} \times 0.035 \text{ L} = 0.0035 \text{ moles}\]

Step 3: Total moles of Cl⁻

Add the moles of Cl⁻ from both sources:\[\text{Total moles of } \mathrm{Cl}^{-} = 0.000625 \text{ moles} + 0.0035 \text{ moles} = 0.004125 \text{ moles}\]

Step 4: Calculate total volume of the solution

Add the volumes of the solutions:\[\text{Total volume} = 25.0 \text{ mL} + 35.0 \text{ mL} = 60.0 \text{ mL} = 0.060 \text{ L} \]

Step 5: Calculate molar concentration of Cl⁻

Use the formula for molarity:\[\text{Molar concentration of } \mathrm{Cl}^{-} = \frac{\text{Total moles of } \mathrm{Cl}^{-}}{\text{Total volume}} = \frac{0.004125 \text{ moles}}{0.060 \text{ L}} = 0.06875 \text{ M}\]

Step 6: Calculate pCl

The pCl is calculated as the negative logarithm (base 10) of the molar concentration of chloride ions:\[\text{pCl} = -\log_{10}(0.06875) \approx 1.16\]

Key Concepts

Chloride Ions

Chloride ions (\(\mathrm{Cl}^{-}\)) are some of the most common ions found in solutions, especially in the context of chemistry experiments and analyses. Chloride ions result from the dissociation of chlorides such as sodium chloride (\(\mathrm{NaCl}\)) and barium chloride (\(\mathrm{BaCl}_{2}\)). In these compounds, chloride acts as an anion and pairs with cations like Na\(^+\) or Ba\(^{2+}\).
In a solution, these compounds dissolve and separate into their ionic components. In the given exercise, knowing how many chloride ions are present after these reactions involves understanding how each compound dissociates:

• \(\mathrm{NaCl}\) dissolves into one \(\mathrm{Na}^+\) and one \(\mathrm{Cl}^{-}\) ion.
• \(\mathrm{BaCl}_{2}\) produces one \(\mathrm{Ba}^{2+}\) and two \(\mathrm{Cl}^{-}\) ions per molecule.

Understanding these details allows one to calculate the exact number of chloride ions in a mixture after combining different solutions, which is crucial for precise solution preparation and analysis.

Solution Preparation

Solution preparation in chemistry often involves mixing various quantities of solutions to achieve a desired concentration of an analyte, like chloride ions in this case. First, it’s important to know the initial concentrations and volumes of each solution being mixed. This exercise focuses on two solutions:

• A 25.0 mL solution of 0.025 M NaCl
• A 35.0 mL solution of 0.050 M BaCl\(_{2}\)

When preparing a solution, make sure to add the volumes together to find the total volume of the mixture. Thus, the total volume here is 60.0 mL or 0.060 L. Mixing solutions while keeping track of the total volume is essential for finding the concentration of specific ions in the combined solution.
Proper measurement and mixing are crucial for ensuring the final solution behaves predictably, especially in reactions or further experiments that require a precise chemical environment.

Molarity Calculation

Molarity is a central concept in chemistry for expressing how concentrated a solution is. It is defined as moles of solute per liter of solution. Calculating molarity involves finding out how many moles of an ion or compound are in your solution, then dividing by the total volume. Here's how it was done in this exercise:
First, calculate the moles of \(\mathrm{Cl}^{-}\) generated by each compound:

• \(\mathrm{NaCl}: 0.025 \text{ M} \times 0.025 \text{ L} = 0.000625 \text{ moles of } \mathrm{Cl}^{-}\)
• \(\mathrm{BaCl}_{2}: 2 \times 0.050 \text{ M} \times 0.035 \text{ L} = 0.0035 \text{ moles of } \mathrm{Cl}^{-}\)

Adding these yields a total of 0.004125 moles of chloride ions. The molarity then becomes:\[\frac{0.004125 \text{ moles}}{0.060 \text{ L}} = 0.06875 \text{ M}\]Calculating molarity is a straightforward but crucial step that impacts the success of chemical reactions and experiments. It's fundamental to ensuring accurate concentration for desired chemical behavior.

pCl Calculation

Understanding how to calculate pCl is important for evaluating the chloride ion concentration on a logarithmic scale, similar to pH.
The formula for calculating the pCl is:\[pCl = -\log_{10}(\text{concentration of } \mathrm{Cl}^{-})\]In our exercise, with a calculated concentration of 0.06875 M for chloride ions, the calculation follows:\[pCl = -\log_{10}(0.06875) \approx 1.16\]The pCl value provides an alternative way to express and understand ion concentration, making it convenient to compare with other logarithmic measures such as pH or pOH. This is particularly useful when dealing with extremely low or high concentrations where logarithmic scales offer a clearer sense of magnitude. Being able to switch between linear and logarithmic expressions of concentration is a valuable skill in chemistry, particularly in analytical contexts.