Question

What is the final volume, in \(\mathrm{mL}\), when \(25.0 \mathrm{~mL}\) of a \(5.00 \mathrm{M}\) \(\mathrm{HCl}\) solution is diluted to each of the following concentrations? a. \(2.50 \mathrm{M} \mathrm{HCl}\) solution b. \(1.00 \mathrm{M} \mathrm{HCl}\) solution c. \(0.500 \mathrm{M} \mathrm{HCl}\) solution

Short answer

a) 50.0 mL, b) 125 mL, c) 250 mL.

Step-by-step solution

- Understanding the Dilution Formula

To solve the problem of dilution, use the formula: \[ C_1 V_1 = C_2 V_2 \] where \( C_1 \) is the initial concentration, \( V_1 \) is the initial volume, \( C_2 \) is the final concentration, and \( V_2 \) is the final volume.

- Plug in the values for part (a)

For part (a), the initial concentration \( C_1 \) is \( 5.00 \mathrm{M} \), the initial volume \( V_1 \) is \( 25.0 \mathrm{~mL} \), and the final concentration \( C_2 \) is \( 2.50 \mathrm{M} \). Plug these values into the formula: \[ 5.00 \times 25.0 = 2.50 \times V_2 \]

- Solve for final volume (V2) for part (a)

Rearrange the equation to solve for \( V_2 \): \[ V_2 = \frac{5.00 \times 25.0}{2.50} \]Calculate the final volume: \[ V_2 = 50.0 \text{ mL} \]

- Plug in the values for part (b)

For part (b), the initial concentration \( C_1 \) is \( 5.00 \mathrm{M} \), the initial volume \( V_1 \) is \( 25.0 \mathrm{~mL} \), and the final concentration \( C_2 \) is \( 1.00 \mathrm{M} \). Plug these values into the formula: \[ 5.00 \times 25.0 = 1.00 \times V_2 \]

- Solve for final volume (V2) for part (b)

Rearrange the equation to solve for \( V_2 \): \[ V_2 = \frac{5.00 \times 25.0}{1.00} \]Calculate the final volume: \[ V_2 = 125 \text{ mL} \]

- Plug in the values for part (c)

For part (c), the initial concentration \( C_1 \) is \( 5.00 \mathrm{M} \), the initial volume \( V_1 \) is \( 25.0 \mathrm{~mL} \), and the final concentration \( C_2 \) is \( 0.500 \mathrm{M} \). Plug these values into the formula: \[ 5.00 \times 25.0 = 0.500 \times V_2 \]

- Solve for final volume (V2) for part (c)

Rearrange the equation to solve for \( V_2 \): \[ V_2 = \frac{5.00 \times 25.0}{0.500} \]Calculate the final volume: \[ V_2 = 250 \text{ mL} \]

Key concepts

Dilution Formula

To understand dilution, you need to learn about the dilution formula. This formula ensures that the amount of solute remains the same before and after dilution. The general form of the equation is: \[ C_1 V_1 = C_2 V_2 \] where:

• \(C_1\): Initial concentration
• \(V_1\): Initial volume
• \(C_2\): Final concentration
• \(V_2\): Final volume

In this formula, as the volume increases, the concentration decreases. This relationship helps to solve various dilution problems efficiently. You just need to plug in the known values and solve for the unknown variable.
For example, in Part (a) of the exercise, we know: \(C_1 = 5.00 \text{ M}\), \( V_1 = 25.0 \text{ mL}\), and \(C_2 = 2.50 \text{ M}\). By plugging these values into the dilution formula, you can find \(V_2\).

Concentration

Concentration describes how much solute is present in a given amount of solvent or solution. It can be measured in various units, but molarity (\(M\)) is the most common in chemistry.
Molarity is defined as the number of moles of solute divided by the volume of the solution in liters (L). In mathematical terms:
\[M = \frac{moles \text{ of solute}}{volume \text{ of solution (in liters)}}\]
High concentration means there are more solute particles in a given volume, while low concentration means fewer solute particles for the same volume. For instance, a \(5.00 \text{ M}\) solution has more solute than a \(0.500 \text{ M}\) solution of the same volume.
When you dilute a solution, you're essentially decreasing the concentration by increasing the volume of the solvent while keeping the amount of solute constant.

Molarity

Molarity is a central concept when dealing with solutions and dilutions. It’s a measure of the concentration of a solute in a solution.
A solution with a molarity of \(1 \text{ M}\) contains one mole of solute per liter of solution. It is important to understand what happens to molarity during dilution. As you add more solvent to a solution, the total volume increases, but the amount of solute remains the same, thereby reducing the molarity.
This is why in Part (b) of the exercise, the final concentration \(C_2\) becomes \(1.00 \text{ M}\) when the initial \(5.00 \text{ M}\) solution is diluted from \(25.0 \text{ mL}\) to \(125 \text{ mL}\).
Remember, molarity is affected by both the amount of solute and the total volume of solution.

Volume Calculation

Calculating volumes during dilution is important to achieving desired concentrations. You often need to find the final volume (\(V_2\)) when the initial concentration (\(C_1\)) and volume (\(V_1\)) are known, along with the final concentration (\(C_2\)). To find the final volume, rearrange the dilution formula:
\[ V_2 = \frac{C_1 V_1}{C_2}\]
Using this equation makes it easy to determine the new volume. For instance, in Part (c) of the exercise, we find \(V_2\) by plugging in \(C_1 = 5.00 \text{ M}\), \(V_1 = 25.0 \text{ mL}\), and \(C_2 = 0.500 \text{ M}\):
\[ V_2 = \frac{5.00 \times 25.0}{0.500} = 250 \text{ mL}\]
This calculation helps to achieve the precise concentration needed in various chemical applications.