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 \(1.000 \mathrm{~L}\). 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 are: stock solution = \(\frac{1.584}{54.94(1.000)}\mathrm{\ M}\), solution A = \(\frac{C1(0.05000)}{1.000}\mathrm{\ M}\), solution B = \(\frac{C1(0.01000)}{0.250}\mathrm{\ M}\), and solution C = \(\frac{C1(0.01000)}{0.500}\mathrm{\ M}\), where \(C1\) is the concentration of the previous solution.

Step-by-step solution

Calculate the moles of manganese metal

To find the number of moles of manganese in the stock solution, we'll use the formula: Moles = mass / molar mass Given: Mass of manganese = 1.584 g Molar mass of manganese = 54.94 g/mol Moles of manganese = \(\frac{1.584\mathrm{\ g}}{54.94\mathrm{\ g/mol}}\)

Calculate the concentration of the stock solution

Now we'll find the concentration of Mn2+ in the stock solution: Concentration = moles / volume Given: Moles of manganese = \(\frac{1.584}{54.94}\mathrm{\ mol}\) Volume = 1.000 L Concentration (stock solution) = \(\frac{1.584}{54.94(1.000)}\mathrm{\ M}\)

Calculate the concentration of solution A

Now we'll use the dilution equation to find the concentration of solution A: \(C1V1=C2V2\) Given: \(C1\) = stock solution concentration \(V1\) = 50.00 mL \(V2\) = 1000.0 mL To make the units consistent, convert mL to L: \(V1 = 50.00\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.05000\mathrm{\ L}\) \(V2 = 1000.0\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 1.000\mathrm{\ L}\) Solve for \(C2\) (concentration of solution A): \(C2 = \frac{C1V1}{V2}\)

Calculate the concentration of solution B

Now we'll repeat the process to find the concentration of solution B, using solution A as our starting solution: Given: \(C1\) = solution A concentration \(V1\) = 10.00 mL \(V2\) = 250.0 mL Convert mL to L: \(V1 = 10.00\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.01000\mathrm{\ L}\) \(V2 = 250.0\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.250\mathrm{\ L}\) Solve for \(C2\)(concentration of solution B): \(C2 = \frac{C1V1}{V2}\)

Calculate the concentration of solution C

Finally, we'll find the concentration of solution C, using solution B as our starting solution: Given: \(C1\) = solution B concentration \(V1\) = 10.00 mL \(V2\) = 500.0 mL Convert mL to L: \(V1 = 10.00\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.01000\mathrm{\ L}\) \(V2 = 500.0\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.500\mathrm{\ L}\) Solve for \(C2\)(concentration of solution C): \(C2 = \frac{C1V1}{V2}\)

Key concepts

Dilution Equation

The dilution equation is a simple and effective way to calculate the concentration of a solution after it has been diluted. It is represented by the formula \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( C_2 \) are the initial and final concentrations of the solution, and \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively.

When diluting a solution, the number of moles of solute remains constant. This means that as you increase the volume of the solution by adding more solvent, the concentration of solute decreases. The dilution equation captures this change efficiently.

Let's walk through an example. Suppose you have a solution with a concentration \( C_1 \) that you want to dilute to a new concentration \( C_2 \). You know the starting volume \( V_1 \) and want to find out what the new volume \( V_2 \) should be, or vice versa. Rearrange the equation to solve for the unknown, creating a flexible tool for various laboratory situations.

• Always ensure volumes are in the same units before using this equation (e.g., convert mL to L).
• Use this equation when you're certain no solute is lost during dilution.
• It's commonly applied in chemistry labs to prepare solutions of desired concentrations by mixing a concentrated solution with a solvent.

Moles Calculation

Understanding moles calculation is crucial in chemistry, as the mole is a fundamental unit used to count atoms and molecules, similar to counting eggs in dozens. To find the number of moles in a sample, use the formula:

\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]

This formula tells you how many entities (atoms, molecules) are present in a given amount of substance by considering the atomic or molecular weight, which is the molar mass, typically in grams per mole (g/mol).

For example, in the given exercise, 1.584 g of manganese metal was dissolved.The molar mass of manganese is 54.94 g/mol. Plug these values into the formula:

\[ \text{Moles of Mn} = \frac{1.584 \text{ g}}{54.94 \text{ g/mol}} \]

Once we determine the moles, we can accurately describe the amount of manganese dissolved. This accuracy is foundational for further calculations, like determining concentration.

Molar Mass

Molar mass is a key concept in chemistry that represents the mass of one mole of a given substance. It's typically expressed in grams per mole (g/mol). Understanding and calculating molar mass is essential for converting between grams and moles.

To find the molar mass, sum the atomic masses of all the atoms in a molecule. For elements, you can simply look up the atomic mass on the periodic table. For compounds, add the atomic masses of the constituent elements, taking into account the number of each type of atom in the formula.

Consider manganese (Mn) as an example, which has a molar mass of 54.94 g/mol. Here's how you might calculate it:

• Locate manganese on the periodic table, noting its atomic mass.
• For compounds, ensure every atom's contribution is included. For example, water (H₂O) involves 2 hydrogen atoms and 1 oxygen atom, summing their atomic masses.

The molar mass is vital for calculations in stoichiometry, where it's used to relate masses of substances to moles, an essential step for reactions and concentration calculations. This skill gives you the ability to measure chemical reactions and create solutions with precise concentrations.