Question

If \(1.00 \mathrm{~mL}\) of a \(2.25 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution needs to be diluted to \(1.00 \mathrm{M}\), what will be its final volume?

Short answer

The final volume is 2.25 mL.

Step-by-step solution

Understand the Problem

We want to dilute a solution of sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) from an initial molarity (\(M_1 = 2.25 \mathrm{M}\)) to a final molarity (\(M_2 = 1.00 \mathrm{M}\)). The goal is to find the final volume (\(V_2\)) when starting with an initial volume (\(V_1 = 1.00 \mathrm{~mL}\)).

Recall the Dilution Formula

The formula for dilutions is given by:\[ M_1 \times V_1 = M_2 \times V_2 \]where \(M_1\) and \(M_2\) are the initial and final molarities, and \(V_1\) and \(V_2\) are the initial and final volumes, respectively.

Plug in the Known Values

Substitute the known values into the dilution equation:\[ 2.25 \times 1.00 = 1.00 \times V_2 \]This equation represents the relationship between the initial and final conditions.

Solve for the Final Volume \(V_2\)

Rearrange the equation to solve for \(V_2\):\[ V_2 = \frac{2.25 \times 1.00}{1.00} \]Calculating this gives:\[ V_2 = 2.25 \mathrm{~mL}\]

Verify the Calculation

Check that the calculations make sense and confirm that the units are consistent: - The units for volume are in milliliters (mL). - The calculation confirms that the final volume needed is 2.25 mL to achieve a 1.00 M solution.

Key concepts

Molarity

Molarity is a fundamental concept in chemistry that helps us understand the concentration of solutions. It's expressed as the number of moles of solute (the substance being dissolved) per liter of solution. This measurement allows chemists to prepare and use solutions with precise concentrations.

• Molarity (\( M \)) is defined by the formula: \[ M = \frac{n}{V} \]where \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters.
• For example, a 2.25 M solution of \( \text{H}_2\text{SO}_4 \) means there are 2.25 moles of sulfuric acid per liter of solution.

Understanding molarity is crucial when performing chemical reactions and dilutions. It ensures that the right amount of substance is present, affecting both the efficiency and outcome of the reaction.
Molarity helps in converting between the volume of a solution and the amount of solute it contains.

Volume Calculation

Calculating the volume in dilution problems is essential to achieve the desired solution concentration. In our exercise, we needed to find the final volume of a diluted sulfuric acid solution.

• The dilution equation used is:\[ M_1 \times V_1 = M_2 \times V_2 \]
• This equation connects initial conditions (\( M_1 \) and \( V_1 \)) to final conditions (\( M_2 \) and \( V_2 \)).

By plugging in the known values, we can solve for the unknown, which is often the final volume. It's a practical method that ensures the solution attains the target concentration.

To arrive at the solution:

• Rearrange the equation to solve for \( V_2 \): \[ V_2 = \frac{M_1 \times V_1}{M_2} \]
• Substitute the given values: \( V_2 = \frac{2.25 \times 1.00}{1.00} = 2.25 \text{ mL} \)

Using the correct units and consistently applying the formula simplifies the process and ensures accuracy.

Chemistry Problem Solving

Chemistry problem solving is a valuable skill. It involves applying principles like molarity and volume calculations to achieve solutions.

• Correctly identifying known values and goals is the first step in solving a problem.
• Choosing the right formula, like the dilution equation, guides the process.

For instance, in our exercise:

• The problem was clear: dilute a concentrated solution to a specific lower molarity.
• By using the dilution formula methodically, we solved for the final volume needed.

Problem-solving in chemistry often requires logical reasoning and an understanding of the underlying scientific principles.
It's not just about substituting into formulas, but also verifying that the solution makes sense in context. Ensuring that units are consistent and calculations are accurate is crucial for obtaining credible results.
This approach builds confidence and competence in handling various quantitative chemistry challenges.