Question

You need to make \(150.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M} \mathrm{NaCl}\) solution. You have solid \(\mathrm{NaCl}\), and your lab partner has a \(2.5 \mathrm{M} \mathrm{NaCl}\) solution. Explain how you each make the \(0.10 \mathrm{M} \mathrm{NaCl}\) solution.

Short answer

To make a 150.0 mL of a 0.10 M NaCl solution using solid NaCl: weigh 0.877 g of solid NaCl, add it to a 150 mL volumetric flask, fill the flask with distilled water up to the 150 mL mark, and stir to dissolve. To make it using the 2.5 M NaCl solution: measure 6 mL of the 2.5 M NaCl solution using a graduated pipette, transfer it to a 150 mL volumetric flask, fill the flask with distilled water up to the 150 mL mark, and stir to ensure even distribution.

Step-by-step solution

Calculate the required moles of NaCl using the M and V of the desired solution

To make a 0.10 M NaCl solution, we need to calculate the moles of NaCl required, using the given molarity and volume information. The formula to find moles is: Moles of solute = Molarity × Volume (in liters) Moles of NaCl = \(0.10 \mathrm{M} \times 0.150 \mathrm{L}\) Moles of NaCl = 0.015 mol Therefore, we need 0.015 moles of NaCl to make 150 mL of a 0.10 M NaCl solution.

Determine the amount of solid NaCl needed

To find out the amount of solid NaCl needed, we must use the molar mass of NaCl, which is 58.44 g/mol. Multiply the needed moles with the molar mass to obtain the mass of NaCl required. Mass of NaCl = Moles of NaCl × Molar Mass of NaCl Mass of NaCl = \(0.015 \mathrm{mol} \times 58.44 \frac{\mathrm{g}}{\mathrm{mol}}\) Mass of NaCl = 0.877 g Thus, 0.877 grams of solid NaCl is required for making the 0.10 M NaCl solution.

Find out the volume of the 2.5 M NaCl solution needed

To use the 2.5 M NaCl solution provided by the lab partner, we need to find the required volume to make a 0.10 M solution. To do this, we would use the formula: \(C_1V_1 = C_2V_2\) Where \(C_1\) and \(V_1\) represent the initial concentration and volume, and \(C_2\) and \(V_2\) represent the final concentration and volume, respectively. Rearranging the formula, we get: \(V_1 = \frac{C_2V_2}{C_1}\) Now, we will plug in the values: \(V_1 = \frac{0.1 \mathrm{M} \times 0.150 \mathrm{L}}{2.5 \mathrm{M}}\) \(V_1 = 0.006 \mathrm{L}\) \(V_1 = 6 \mathrm{mL}\) Hence, 6 mL of the 2.5 M NaCl solution is needed.

Making the 0.10 M NaCl solution using both methods

For Method 1, using solid NaCl: 1. Weigh 0.877 grams of solid NaCl. 2. Add the solid NaCl to a 150 mL volumetric flask. 3. Fill the flask with distilled water until the 150 mL mark is reached. 4. Stir the solution to ensure that the NaCl is completely dissolved to obtain a 0.10 M NaCl solution. For Method 2, using the 2.5 M NaCl solution: 1. Use a graduated pipette to measure 6 mL of the 2.5 M NaCl solution. 2. Transfer the 6 mL of the NaCl solution to a 150 mL volumetric flask. 3. Fill the flask with distilled water until the 150 mL mark is reached. 4. Stir the solution to ensure the NaCl solution is evenly distributed to make a 0.10 M NaCl solution.

Key concepts

Solution Preparation

When preparing a solution, the goal is to dissolve the desired amount of solute (which could be a solid or a concentrated liquid solution) in a solvent, typically water, to achieve a specific concentration. In the context of the provided exercise, two methods were used to prepare a 0.10 M NaCl solution. The first method involves dissolving a calculated mass of solid NaCl. The second method involves diluting an existing more concentrated NaCl solution.

• For the first method, weigh the solid NaCl and add it to a container with less solvent than the total desired volume. Once dissolved thoroughly, additional solvent is added to reach the final desired volume.
• For the second method, measure a specific volume of a more concentrated NaCl solution. This is diluted by adding more solvent until the total desired volume is achieved, lowering the solution's concentration.

Both techniques involve achieving the same final concentration through careful measurement and mixing.

Dilution

In chemistry, dilution is the process of decreasing the concentration of a solute in a solution, usually by adding more solvent. It is a critical technique often used to achieve a desired molarity from a more concentrated stock solution.
To calculate how much of a concentrated solution is needed for dilution, the formula \( C_1V_1 = C_2V_2 \) is used:

• \(C_1\): Initial concentration of the stock solution.
• \(V_1\): Volume of the stock solution needed.
• \(C_2\): Desired concentration after dilution.
• \(V_2\): Final total volume of the solution.

By rearranging the formula, you can solve for \(V_1\), which gives the volume of the more concentrated solution to be used. This is then mixed with solvent to reach the desired concentration and volume.

Moles Calculation

Calculating the moles of a particular substance helps in understanding how much of the substance is there present in a given mass or volume, useful in preparing solutions. The formula \( \text{Moles} = \text{Molarity} \times \text{Volume (in liters)} \) allows for the calculation of moles based on the molarity and volume of a solution.

• Molarity (\(M\)) is the number of moles of solute per liter of solution.
• Volume must be converted to liters from milliliters if necessary.

Thus, multiplying the molarity by the volume gives the total moles of the solute necessary to achieve a desired solution concentration. This concept is essential across various chemistry tasks, whether measuring solids or handling liquids.

Concentration

Concentration refers to the amount of solute present in a given volume of solution, typically expressed in terms of molarity (\(M\)). Molarity is a straightforward way to express concentration, allowing chemists to communicate how potent a solution is.

• Molarity (\(M\)) is expressed as moles of solute per liter of solution: \(M = \frac{\text{moles of solute}}{\text{liters of solution}}\).
• Knowing a solution's concentration helps determine how it will react, its potency, and how much is required for further processes or experiments.

In scenarios where the concentration needs adjustment, understanding dilution or the addition of further solute can modify a solution precisely, tailoring it to specific needs. Concentration is a central concept in chemistry, playing a role in reactions, experiments, and product formulations.