Question

A stock solution containing \(\mathrm{Mn}^{2+}\) ions was prepared by dissolving 1.584 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions were then prepared by dilution: For solution \(A, 50.00 \mathrm{mL}\) of stock solution was diluted to \(1000.0 \mathrm{mL}\) For solution \(B, 10.00 \mathrm{mL}\) of solution \(A\) was diluted to \(250.0 \mathrm{mL}\). For solution \(C, 10.00 \mathrm{mL}\) of solution \(B\) was diluted to \(500.0 \mathrm{mL}\). Calculate the concentrations of the stock solution and solutions \(A, B,\) and \(C .\)

Short answer

The concentrations of the stock solution and solutions A, B, and C are as follows: - Stock solution: 0.0288 M - Solution A: 0.00144 M - Solution B: \(5.76 \times 10^{-5}\) M - Solution C: \(1.152 \times 10^{-5}\) M

Step-by-step solution

Calculate the number of moles of manganese in the stock solution.

To calculate the number of moles of Mn2+ ions in the stock solution, we can use the formula: number of moles = mass / molar mass First, we need to find the molar mass of manganese (Mn). From the periodic table, the molar mass of Mn is 54.94 g/mol. So: number of moles = 1.584g / 54.94 g/mol = 0.0288 mol

Calculate the concentration of the stock solution.

Now that we have the number of moles of Mn2+ ions, we can calculate the concentration of the stock solution using the formula: Concentration = number of moles / volume The volume of the stock solution is 1.000 L. So: Concentration = 0.0288 mol / 1.000 L = 0.0288 mol/L The concentration of the stock solution is 0.0288 M.

Calculate the concentration of solution A.

To find the concentration of solution A, we can use the dilution formula: C1 * V1 = C2 * V2 where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the concentration and volume of solution A. We know that 50.00 mL of the stock solution was diluted to 1000 mL. Plugging in the values: (0.0288 mol/L) * (50.00 mL) = C2 * (1000.0 mL) Now solve for C2: C2 = (0.0288 mol/L) * (50.00 mL) / (1000.0 mL) = 0.00144 mol/L The concentration of solution A is 0.00144 M.

Calculate the concentration of solution B.

Using the same dilution formula, we can calculate the concentration of solution B: C1 * V1 = C2 * V2 Here, C1 and V1 are the concentration and volume of solution A, and C2 and V2 are the concentration and volume of solution B. We know that 10.00 mL of solution A was diluted to 250.0 mL. Plugging in the values: (0.00144 mol/L) * (10.00 mL) = C2 * (250.0 mL) Now solve for C2: C2 = (0.00144 mol/L) * (10.00 mL) / (250.0 mL) = 5.76 * 10^(-5) mol/L The concentration of solution B is 5.76 * 10^(-5) M.

Calculate the concentration of solution C.

Finally, we can calculate the concentration of Solution C using the dilution formula: C1 * V1 = C2 * V2 Here, C1 and V1 are the concentration and volume of solution B, and C2 and V2 are the concentration and volume of solution C. We know that 10.00 mL of solution B was diluted to 500.0 mL. Plugging in the values: (5.76 * 10^(-5) mol/L) * (10.00 mL) = C2 * (500.0 mL) Now solve for C2: C2 = (5.76 * 10^(-5) mol/L) * (10.00 mL) / (500.0 mL) = 1.152 * 10^(-5) mol/L The concentration of solution C is 1.152 * 10^(-5) M.

Key concepts

Dilution Calculation

Dilution is a crucial concept in chemistry that allows us to reduce the concentration of a solution by adding more solvent. The key formula used in dilution calculations is:\[ C_1 \times V_1 = C_2 \times V_2 \]Here:

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

Using this equation, you can determine how the concentration changes as you adjust the volume. In our exercise, solutions A, B, and C were prepared by sequentially diluting a stock solution. For each step:

• First, we calculate the number of moles in the smaller volume (\( V_1 \)).
• Then, divide by the new total volume (\( V_2 \)) to find the new concentration (\( C_2 \)).

Understanding this process helps predict the concentrations at each stage, making it easier to plan experiments or processes that require specific solution strengths.

Mole Concept

The mole concept is fundamental in chemistry, allowing us to quantify amounts in chemical reactions. A mole represents \( 6.022 \times 10^{23} \) entities (Avogadro's number) of any substance.In the context of our problem, we first determined how many moles of manganese were present in the stock solution. This involves the formula:\[ \text{number of moles} = \frac{\text{mass}}{\text{molar mass}} \]For manganese, we used its molar mass found in the periodic table, which is \( 54.94 \text{ g/mol} \). By dividing the mass of manganese used \( (1.584 \text{ g}) \) by its molar mass, we found the number of moles.This foundational step is essential because the number of moles will determine the initial concentration in the stock solution and guide subsequent dilution calculations. It tells us how much substance is actually present, regardless of any chemical formula, making it a versatile tool for chemists.

Molarity Determination

Molarity is a measure of the concentration of a solution, expressed in moles of solute per liter of solution. In our problem, we calculated the molarity of the initial stock solution and then proceeded with solutions A, B, and C.Start with the formula:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]For the stock solution, we used the moles calculated earlier and divided by the total volume (1.000 L). This gave us the molarity of the starting solution. Each subsequent dilution followed the same principle but adjusted for the new volumes.Molarity is significant because it allows chemists to easily relate volumes and concentrations in a practical way, crucial for mixing solutions accurately. This concept underpins many aspects of laboratory work, providing a clear measure of how solute is distributed in a solution.