Problem 54 - Chemistry and Chemical Reactivity

Question

Which of the following methods would you use to prepare \(300 . \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\) (a) Add \(30.0 \mathrm{mL}\) of \(1.50 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to \(270 .\) mL of water. (b) Dilute \(250 .\) mL of \(0.600 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a volume of \(300 . \mathrm{mL}\).

Step-by-Step Solution

Answer

Use method (b): dilute 250 mL of 0.600 M solution to 300 mL.

Step 1: Determine the moles needed

Calculate the number of moles of \( \text{K}_2 \text{Cr}_2 \text{O}_7 \) needed using the concentration and volume formula: \( \text{Moles} = \text{Molarity} \times \text{Volume} \). Here, the molarity is 0.500 M and the volume is 0.300 L (since 300 mL = 0.300 L). Thus, \( \text{Moles} = 0.500 \times 0.300 = 0.150 \text{ moles} \).

Step 2: Calculate moles from Option (a)

From Option (a), calculate the moles of \( \text{K}_2 \text{Cr}_2 \text{O}_7 \) by using the formula \( \text{Moles} = \text{Molarity} \times \text{Volume} \). The given molarity is 1.50 M and the volume is 0.030 L (since 30.0 mL = 0.030 L). Thus, \( \text{Moles} = 1.50 \times 0.030 = 0.045 \text{ moles} \), which is not enough.

Step 3: Calculate moles from Option (b)

For Option (b), we use the dilution formula \( M_1 \times V_1 = M_2 \times V_2 \). Here, \( M_1 = 0.600 \), \( V_1 = 0.250 \), \( M_2 = 0.500 \), and \( V_2 = 0.300 \). Calculate \( M_1 \times V_1 = 0.600 \times 0.250 = 0.150 \text{ moles} \), which matches the required moles.

Step 4: Conclusion on correct option

Compare the moles needed with the moles calculated in each option. Only Option (b) provides the exact moles needed (0.150 moles) for the desired solution of \( 300 \text{ mL} \) at \( 0.500 M \). Therefore, the correct method to prepare the solution is by diluting 250 mL of 0.600 M solution to a final volume of 300 mL.

Key Concepts

Molarity

Molarity is a measure of how concentrated a solution is. It tells us the number of moles of a solute present in one liter of solution. Understanding molarity is crucial in solution preparation because it helps us quantify the concentration of a particular chemical in a given volume.
To calculate molarity, you use the formula:

• Molarity (M) = Moles of solute / Volume of solution in liters

This is why molarity is often expressed in units of moles per liter (mol/L). Knowing the molarity allows chemists to prepare solutions of desired concentrations by determining the amount of solute required and the volume of solvent to add. When preparing laboratory solutions, accuracy in molarity ensures that experiments and reactions proceed with the correct chemical concentrations, leading to reliable results.

Dilution formula

The dilution formula is a handy tool that helps adjust the concentration of a solution by adding more solvent. This is particularly useful when you have a solution of known higher concentration and need to produce a larger volume of a lower concentration.
The dilution formula is expressed as:

• \( M_1 \times V_1 = M_2 \times V_2 \)

Where:

• \( M_1 \) is the initial molarity,
• \( V_1 \) is the initial volume,
• \( M_2 \) is the final molarity,
• \( V_2 \) is the final volume.

This formula assumes that volume additions do not compress the solution. By rearranging the equation, you can find any unknown value if the others are provided. For example, if you know you have a 0.600 M solution and want to make a 0.500 M solution, you can calculate the volumes needed to achieve this new concentration. It precisely led to finding that diluting 250 mL of 0.600 M solution gives you 300 mL of a 0.500 M solution, as seen in the solution steps.

Chemical Concentration

Chemical concentration refers to the amount of a substance, known as a solute, present in a given quantity of solvent or solution. This is a key concept in chemistry and is essential for preparing solutions correctly for various applications, including chemical reactions and laboratory experiments.
Concentration can be expressed in several ways, with molarity being one of the most common. It indicates how much solute is present in a specific volume of solution. When working with chemical solutions, it is important to accurately measure and calculate concentrations to ensure that reactions occur as expected. For instance, the desired solution of 0.500 M was needed in a volume of 300 mL for the exercise. Choosing the correct concentration and volume is essential not only for stoichiometric considerations but also for safety and efficacy in chemical processes.

Moles calculation

Calculating moles is an essential skill in chemistry because moles are a fundamental unit that measures the amount of substance. To calculate moles in solution preparation, you'll often use the formula:

• Moles = Molarity \( \times \) Volume

This calculation is straightforward. First, convert the volume from milliliters to liters (since molarity is in moles per liter). Then, multiply the molarity by the volume to determine the number of moles of the solute. In the exercise, calculating moles allowed us to compare different preparation methods to ensure the correct amount of solute was used and adjust as needed.
Understanding moles calculation allows chemists to precisely control how much of a compound is present in a reaction or solution, critical for making informed decisions and accurate predictions in their experiments. By mastering mole calculations, you ensure that the solutions prepared are neither too concentrated nor too dilute for the desired application.