Problem 25 - Chemistry Matter and Change

Question

How many milliliters of a 5.0 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) stock solution would you need to prepare 100.0 \(\mathrm{mL}\) of 0.25 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Step-by-Step Solution

Answer

To prepare 100.0 mL of 0.25 M H₂SO₄ solution using a 5.0 M H₂SO₄ stock solution, you would need \(5.0 \mathrm{mL}\) of the stock solution. This is calculated using the dilution formula: \(M_1V_1 = M_2V_2\).

Step 1: Identify the dilution formula

The dilution formula is as follows: \[M_1V_1 = M_2V_2\] where \(M_1\) is the initial concentration of the stock solution, \(V_1\) is the volume of the stock solution, \(M_2\) is the desired final concentration, and \(V_2\) is the desired final volume.

Step 2: Identify given information

We are given the following information: - Initial concentration of H₂SO₄ stock solution (\(M_1\)): 5.0 M - Desired final concentration of H₂SO₄ solution (\(M_2\)): 0.25 M - Desired final volume of H₂SO₄ solution (\(V_2\)): 100.0 mL

Step 3: Plug the given values into the dilution formula

Now, we plug in the given values for \(M_1\), \(M_2\), and \(V_2\) into the dilution formula: \[(5.0 \mathrm{M})V_1 = (0.25 \mathrm{M})(100.0 \mathrm{mL})\]

Step 4: Solve for the volume of stock solution (\(V_1\))

To find \(V_1\), we simply divide both sides by \(M_1\): \[V_1 = \frac{(0.25 \mathrm{M})(100.0 \mathrm{mL})}{5.0 \mathrm{M}}\] Now, let's calculate the value of \(V_1\):

Step 5: Final answer

\[V_1 = \frac{(0.25 \mathrm{M})(100.0 \mathrm{mL})}{5.0 \mathrm{M}} = 5.0 \mathrm{mL}\] Therefore, you would need 5.0 mL of the 5.0 M H₂SO₄ stock solution to prepare 100.0 mL of 0.25 M H₂SO₄ solution.

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula for molarity is:

• \[ M = \frac{n}{V} \]

where \( M \) is the molarity, \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters. Understanding molarity is crucial as it helps in determining how much solute is present in a given volume of solution.
In the exercise, the stock solution of sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4} \)) has a molarity of 5.0 M, indicating a high concentration of solute. The goal was to prepare a more dilute 0.25 M solution, which requires dilution.

Solution Preparation

Solution preparation involves the process of mixing solutes and solvents to achieve a desired concentration. The key steps in preparing a solution include:

• Calculating the amount of solute or stock solution needed using the dilution formula: \[ M_1V_1 = M_2V_2 \]
• Mixing the calculated amount of concentrated solution with a solvent (usually water) to reach the final desired volume and concentration.

By understanding these steps, students can accurately prepare solutions for various laboratory and practical purposes.
In our example, the target was to prepare 100.0 mL of a 0.25 M solution from a 5.0 M stock. Using the dilution formula ensures the correct proportions of stock to solvent are used.

Volume Calculation

Volume calculation is crucial when it comes to preparing solutions, especially during dilution. This involves determining how much of the concentrated stock solution must be used to make a new solution of desired concentration and volume.
For this, the dilution formula \( M_1V_1 = M_2V_2 \) is applied. In our exercise, we needed to find \( V_1 \), the volume of the stock solution needed:

• Re-arranged formula: \[ V_1 = \frac{M_2V_2}{M_1} \]
• Substitute the values and solve: \[ V_1 = \frac{(0.25 \mathrm{M})(100.0 \mathrm{mL})}{5.0 \mathrm{M}} \]
• Calculate: \[ V_1 = 5.0 \mathrm{mL} \]

This calculation tells us that 5.0 mL of the 5.0 M stock solution is needed to prepare 100.0 mL of a 0.25 M solution. Understanding these computations aids students in carrying out solution preparations correctly.