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 mL of stock solution was diluted to 1000.0 mL. For solution B, 10.00 mL of solution A was diluted to 250.0 mL. For solution C, 10.00 mL of solution B was diluted to 500.0 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 \(2.88 \times 10^{-2} \mathrm{M}\), \(1.44 \times 10^{-3} \mathrm{M}\), \(5.76 \times 10^{-5} \mathrm{M}\), and \(1.15 \times 10^{-5} \mathrm{M}\), respectively.

Step-by-step solution

Find the moles of Mn in the stock solution

First, we need to find the moles of manganese metal by dividing the given mass (1.584 g) by its molar mass (54.94 g/mol): Moles of Mn = \(\frac{1.584 \mathrm{g}}{54.94 \mathrm{g/mol}}\) Moles of Mn = \(2.88 \times 10^{-2} \mathrm{mol}\)

Calculate the concentration of the stock solution

Now, we'll find the molar concentration of the Mn^2+ ions in the stock solution by dividing the moles of Mn by the final volume (1.000 L): Concentration of stock solution = \(\frac{2.88 \times 10^{-2} \mathrm{mol}}{1.000 \mathrm{L}}\) Concentration of stock solution = \(2.88 \times 10^{-2} \mathrm{M}\)

Calculate the concentration of solution A

To calculate the concentration of solution A, we'll use the dilution formula: C1V1 = C2V2 \((2.88 \times 10^{-2} \mathrm{M})(50.00 \times 10^{-3} \mathrm{L}) = (C_A)(1000.0 \times 10^{-3} \mathrm{L})\) Solving for C_A, we get: C_A = \(1.44 \times 10^{-3} \mathrm{M}\)

Calculate the concentration of solution B

Using the same dilution formula, we'll find the concentration of solution B: \((1.44 \times 10^{-3} \mathrm{M})(10.00 \times 10^{-3} \mathrm{L}) = (C_B)(250.0 \times 10^{-3} \mathrm{L})\) Solving for C_B, we get: C_B = \(5.76 \times 10^{-5} \mathrm{M}\)

Calculate the concentration of solution C

Finally, we'll find the concentration of solution C: \((5.76 \times 10^{-5} \mathrm{M})(10.00 \times 10^{-3} \mathrm{L}) = (C_C)(500.0 \times 10^{-3} \mathrm{L})\) Solving for C_C, we get: C_C = \(1.15 \times 10^{-5} \mathrm{M}\) So, the concentrations of the stock solution and solutions A, B, and C are \(2.88 \times 10^{-2} \mathrm{M}\), \(1.44 \times 10^{-3} \mathrm{M}\), \(5.76 \times 10^{-5} \mathrm{M}\), and \(1.15 \times 10^{-5} \mathrm{M}\), respectively.

Key concepts

Molar Concentration

Molar concentration, also known as molarity, measures the amount of a solute that is dissolved in a given volume of solution. It is expressed in moles per liter (M). In the context of a solution, molarity helps to understand how densely the solute particles are packed within the solution.

To calculate molar concentration, you take the number of moles of the solute and divide it by the volume of the solution in liters. Here, the molar concentration of the \(\mathrm{Mn^{2+}}\) ions in the stock solution is calculated by using the formula:

\[ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
Understanding molarity is crucial for ensuring precision when mixing solutions in labs and for predicting reaction outcomes.

Dilution Formula

The dilution formula is a powerful tool used to determine the concentration of a solution that has been diluted. This means you have added more solvent to a solution, which decreases its concentration.

This formula is expressed as:

• \( C_1 V_1 = C_2 V_2 \)

Here, \( C_1 \) and \( V_1 \) are the concentration and volume of the initial solution, while \( C_2 \) and \( V_2 \) are the concentration and volume after dilution.

This equation works based on the principle that the amount of solute (in moles) remains constant before and after dilution. It's a valuable equation for calculating final concentrations when more solvent is added. A simple rearrangement allows us to find the unknown concentration after dilution.

Mn2+ Ions

Manganese ions, specifically \(\mathrm{Mn^{2+}}\), are an essential part of this exercise. When manganese metal dissolves, it loses electrons to form these ions. These positive ions are what give the solution its conductive properties.

Understanding ion formation is important in chemistry since it influences many solution characteristics. Especially in analytical chemistry, recognizing the role of such ions is pivotal when performing titrations and other quantitative measurements.

In the given exercise, the concentration of manganese ions effectively measures how many positively charged particles are floating in our solutions post-dilution.

Moles Calculation

Calculating moles is a fundamental step in determining the molarity and involves converting mass into moles using the substance's molar mass. The molar mass of manganese is approximately 54.94 g/mol.

The formula used in this calculation is:

• \( \text{Moles} = \frac{\text{Mass of substance (g)}}{\text{Molar mass (g/mol)}} \)

This allows chemists to work with a measureable weight and relate it to the number of atoms or molecules.

This step is crucial in labs to predict how substances will react and helps in creating solutions of desired concentrations through precise calculation. Each step in calculating moles brings us closer to comprehending the behavior of chemicals in various conditions.