Problem 49 - Chemistry and Chemical Reactivity

Question

Which of the following methods would you use to prepare \(1.00 \mathrm{L}\) of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) Dilute \(20.8 \mathrm{mL}\) of \(6.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) to a volume of \(1.00 \mathrm{L}\) (b) Add \(950 .\) mL. of water to 50.0 mL of \(3.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\)

Step-by-Step Solution

Answer

Use method (a) to dilute 20.8 mL of 6.00 M H₂SO₄ to 1.00 L.

Step 1: Understand the Problem

We need to prepare 1.00 L of 0.125 M H₂SO₄ solution. We must decide whether to use method (a) or (b) by checking which dilution results in the desired molarity.

Step 2: Calculate Moles for Desired Solution

The desired solution is 1.00 L of 0.125 M H₂SO₄. To find the moles required, use the molarity formula:\[ C = \frac{n}{V} \] Where \( C = 0.125 \text{ M} \), and \( V = 1.00 \text{ L} \), so the moles \( n \) is:\[ n = 0.125 \times 1.00 = 0.125 \text{ moles} \]

Step 3: Analyze Option (a)

For option (a), use the dilution formula:\[ C_1V_1 = C_2V_2 \]Where \( C_1 = 6.00 \text{ M} \), \( V_1 \) is the volume of the concentrated solution, \( C_2 = 0.125 \text{ M} \), and \( V_2 = 1.00 \text{ L} \). Solve for \( V_1 \):\[ 6.00 \times V_1 = 0.125 \times 1.00 \]\[ V_1 = \frac{0.125}{6.00} \approx 0.0208 \text{ L} \approx 20.8 \text{ mL} \]

Step 4: Analyze Option (b)

For option (b), the solution composition would be:- Initial moles in 50.0 mL of 3.00 M:\[ n = C \times V = 3.00 \times 0.0500 = 0.150 \text{ moles} \]- Dilute this with 950 mL, resulting in total volume of\[ 1.00 \text{ L} (50.0 \text{ mL} + 950 \text{ mL}) \]- Final concentration \( C \):\[ C = \frac{0.150}{1.00} = 0.150 \text{ M} \]

Step 5: Choose the Correct Method

Option (a) yields a concentration of 0.125 M, which matches the desired concentration. Option (b) results in a 0.150 M solution, which is too high. Therefore, option (a) is the correct method to use.

Key Concepts

Molarity

Molarity is a fundamental concept in chemistry that describes the concentration of a solute in a solution. It is expressed as the number of moles of solute per liter of solution. This is why it is sometimes called "molar concentration." The formula for molarity is \[ C = \frac{n}{V} \]where

• \(C\) represents molarity in moles per liter (M).
• \(n\) is the number of moles of solute.
• \(V\) is the volume of solution in liters.

Understanding molarity is crucial in solution preparation and lab work because it allows chemists to precisely control the nature of a chemical reaction.
For example, controlling the molarity of a solution ensures that the chemical reactions involved are not too slow or too fast.
Calculating molarity helps in determining precisely how much of a solute is needed to achieve a desired solution concentration.

Solution Preparation

Preparation of a chemical solution involves several key steps to ensure accuracy and consistency in experiments. One primary method is dilution, especially when working with highly concentrated solutions.
When preparing a dilute solution from a concentrated stock, you use the dilution formula:\[ C_1V_1 = C_2V_2 \]Here,

• \(C_1\) and \(C_2\) are the initial and final concentrations, respectively.
• \(V_1\) and \(V_2\) are the initial and final volumes.

This approach helps in safely preparing solutions for experiments without directly working with large quantities of highly concentrated substances.
For instance, in our exercise, method (a) required taking 20.8 mL of 6.00 M \(\text{H}_2\text{SO}_4\) and diluting it to reach 1.00 L of a 0.125 M solution.
This method is preferred for accuracy and managing the reactivity of the substances involved.

Concentration Calculation

In chemistry, calculating concentration is vital in determining how a reactant will behave or respond in a solution.
It can involve measuring the amount of solute in a given volume, which is essential in both experimental and industrial chemistry.
The exercise illustrates two main calculation steps:

• Calculating the moles needed for the desired solution using molarity: \( n = C \times V \).
• Applying the required volume calculations using the dilution formula: \( C_1V_1 = C_2V_2 \).

By accurately calculating concentrations, chemists can predict the outcome and efficiency of reactions. This ensures that the reactants are mixed in appropriate ratios.
In practical terms, this helps to avoid overuse of chemicals and ensures the safety and effectiveness of laboratory procedures.
Correct concentration calculations, like in option (a) of the exercise, lead to the preparation of solutions with the desired molarity, necessary for reliable experiment results.