Question

What volume of 0.25 M HCl solution must be diluted to prepare 1.00 L of 0.040 M HCl?

Short answer

To prepare 1.00 L of 0.040 M HCl from a 0.25 M HCl solution, you need to dilute 160 mL of the 0.25 M HCl solution.

Step-by-step solution

Identify the given information

From the exercise, we know that: - Initial concentration (\(C_1\)) = 0.25 M - Final concentration (\(C_2\)) = 0.040 M - Final volume (\(V_2\)) = 1.00 L

Use the dilution formula to solve for the initial volume

Since we have all the values except for the initial volume (\(V_1\)), we can use the dilution formula: \(C_1 \times V_1 = C_2 \times V_2\) Plug in the given values: \(0.25 \times V_1 = 0.040 \times 1.00\)

Solve for the initial volume

Now, we need to solve for \(V_1\): \(V_1 = \frac{0.040 \times 1.00}{0.25}\) Divide and simplify: \(V_1 = 0.16\)

Convert the initial volume to the appropriate units

The initial volume is in liters, but it is more convenient to express it in milliliters. To convert, we can use the following conversion factor: 1 L = 1000 mL So, multiply the initial volume by 1000: \(V_1 = 0.16 \times 1,000\) Which gives: \(V_1 = 160\text{ mL}\) Now we know that \(160\) mL of 0.25 M HCl solution must be diluted to prepare 1.00 L of 0.040 M HCl.

Key concepts

Understanding Molarity

Molarity is a key concept when dealing with solutions, especially in chemistry. It measures the concentration of a solution. Specifically, it is defined as the number of moles of solute (the substance being dissolved) per liter of solution. This can be expressed as:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]Consider molarity as the concentration strength of your solution. A high molarity value indicates a strong, concentrated solution, whereas a low molarity indicates a more diluted solution.**Why Molarity Matters:**

• It helps in comparing concentrations of different solutions.
• It's crucial for chemical reactions where stoichiometry depends on the reactants' concentrations.
• It allows chemists to predict the outcomes of mixing different solutions.

In our exercise, molarity is vital as it guides us on how concentrated the HCl solution needs to be both initially and finally, which helps in the dilution process.

Solution Preparation Techniques

Preparing a solution requires precision and accuracy. Follow these steps to ensure a properly prepared solution which can be crucial for experiments and reactions: **Steps in Solution Preparation:**

• **Measure Solute:** Start with accurately measuring the amount of solute needed. This is often determined using the molarity and desired volume of your solution.
• **Dissolve Solute:** Next, dissolve it in a small volume of solvent. Stirring or gentle heating might help facilitate this.
• **Dilute to Volume:** Finally, add more solvent until the total volume reaches the desired amount.

In our exercise, we dealt with the preparation of a diluted solution from a more concentrated hydrochloric acid (HCl). Starting with the correct volume of a 0.25 M solution, it was diluted to achieve a less concentrated 0.040 M solution in a total volume of 1.00 L. Careful preparation ensures the solution meets the desired molarity and volume exactly.

Essential Dilution Formula

Dilution is a process where a solution's concentration is reduced by adding more solvent. The underlying principle of dilution is captured by the dilution formula:\[ C_1 \times V_1 = C_2 \times V_2 \]Where:

• \( C_1 \) is the initial concentration.
• \( V_1 \) is the initial volume.
• \( C_2 \) is the final concentration.
• \( V_2 \) is the final volume.

This formula ensures that the amount of solute remains constant before and after adding the solvent. The change occurs in the concentration alone, as the solvent increases, expanding the volume.**Using the Dilution Formula:**

• Determine the initial concentration and volume while knowing the target final concentration and volume.
• Plug these values into the formula to solve for the unknown, typically the initial volume or final concentration.

In the given exercise, by using this formula, we were able to determine that 160 mL of the 0.25 M HCl needed to be diluted to achieve the 1.00 L of 0.040 M solution. This calculation is central to accurately executing the dilution process.