Problem 111 - General Chemistry

Question

A stock solution of potassium dichromate, \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), is made by dissolving \(89.3 \mathrm{~g}\) of the compound in \(1.00 \mathrm{~L}\) of solution. How many milliliters of this solution are required to prepare \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\)

Step-by-Step Solution

Answer

329.4 mL of stock solution is needed.

Step 1: Calculate Molar Mass

To determine how much of the stock solution is needed, we first need to calculate the molar mass of potassium dichromate, \(\mathrm{K}_{2} \mathrm{Cr}_{2}\mathrm{O}_{7}\). Adding up the atomic masses, we get:- Potassium (K): \(39.1 \times 2 = 78.2 \, \text{g/mol}\)- Chromium (Cr): \(51.9961 \times 2 = 104 \text{g/mol}\)- Oxygen (O): \(16 \times 7 = 112 \, \text{g/mol}\)Total molar mass = \(78.2 + 104 + 112 = 294.2 \, \text{g/mol}\).

Step 2: Calculate Molarity of Stock Solution

The molarity (concentration) of the stock solution is calculated by dividing the mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2}\mathrm{O}_{7}\) by its molar mass and then dividing by the volume of the solution in liters. \[\text{Molarity} = \frac{89.3 \, \text{g}}{294.2 \, \text{g/mol} \times 1.00 \, \text{L}} = 0.3036 \, \text{mol/L}.\]

Step 3: Use Dilution Formula

To find the volume of the stock solution needed to prepare \(1.00 \, \text{L}\) of \(0.100 \, \text{M}\) solution, we use the dilution formula \(M_1V_1 = M_2V_2\). Here, \(M_1\) is the molarity of the stock solution, \(V_1\) is the volume of the stock solution, \(M_2\) is the desired molarity, and \(V_2\) is the desired volume.\[0.3036 \, \text{mol/L} \times V_1 = 0.100 \, \text{mol/L} \times 1.00 \, \text{L}\]

Step 4: Solve for V_1

Rearrange the formula from the previous step to solve for \(V_1\).\[V_1 = \frac{0.100 \, \text{mol/L} \times 1.00 \, \text{L}}{0.3036 \, \text{mol/L}} = 0.3294 \, \text{L}.\] Convert \(0.3294 \, \text{L}\) to milliliters by multiplying by \(1000\), getting \(329.4 \, \text{mL}\).

Step 5: Conclusion: Determine Volume in mL

The stock solution required to prepare \(1.00 \, \text{L}\) of \(0.100 \, \text{M}\) \(\mathrm{K}_{2} \mathrm{Cr}_{2}\mathrm{O}_{7}\) is \(329.4 \, \text{mL}\).

Key Concepts

Molarity Calculation

Molarity is a measure of concentration, indicating how much solute is present in a specific volume of solution. It is expressed in moles per liter (mol/L). To calculate molarity, you need to know two things: the number of moles of solute and the volume of the solution in liters. This formula is helpful when preparing solutions or performing dilution calculations.

The general formula for molarity (M) is:

• Molarity \( M \) = \( \frac{\text{moles of solute}}{\text{liters of solution}} \)

In our exercise, we calculated the molarity of the stock potassium dichromate solution by dividing the mass of the compound by its molar mass to get moles, and then dividing by the solution's volume in liters. This gives a clear understanding of how concentrated the solution is, which is crucial for further dilution calculations.

Knowing how to calculate molarity allows you to precisely prepare solutions with desired concentrations, a skill that's commonly used in laboratory settings.

Potassium Dichromate Solution

Potassium dichromate, \( \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \), is an orange-red crystalline solid commonly used in laboratories. It can act as a powerful oxidizing agent, making it valuable for various chemical reactions, particularly in stoichiometry. To prepare solutions of potassium dichromate, it is crucial to accurately calculate and measure to ensure the desired concentration and volume.

When making a potassium dichromate solution like in our exercise, precise weighing of the chemical is fundamental. Once the compound is dissolved into a specific volume of water, it forms a homogeneous solution. This process confirms that each portion of the solution shares an equal concentration of potassium dichromate.

In laboratory settings, simplicity and accuracy are critical. Ensuring all measurements are precise helps in maintaining the efficiency and reliability of experimental results when using the solution as a reagent.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It provides a mathematical framework to predict the outcomes of chemical reactions and is essential for calculating the amounts of substances needed or produced in a given reaction.

In the context of our exercise, stoichiometry plays a role in determining how much of the stock potassium dichromate solution is needed to achieve a specific concentration and volume. This involves using the dilution formula \( M_1V_1 = M_2V_2 \). Here, \( M_1 \) and \( V_1 \) represent the molarity and volume of the stock solution, whereas \( M_2 \) and \( V_2 \) represent the desired molarity and volume of the diluted solution.

Understanding stoichiometry is essential for planning and conducting experiments. It ensures that reactants are measured accurately to achieve the desired reaction outcomes, preventing waste and ensuring safety by avoiding the use of excess chemicals.