Question

Determine the volume, in milliliters, required to prepare each of the following diluted solutions: a. \(255 \mathrm{~mL}\) of a \(0.200 \mathrm{M} \mathrm{HNO}_{3}\) solution from a \(4.00 \mathrm{M} \mathrm{HNO}_{3}\) solution b. \(715 \mathrm{~mL}\) of a \(0.100 \mathrm{M} \mathrm{MgCl}_{2}\) solution using a \(6.00 \mathrm{M}\) \(\mathrm{Mg} \mathrm{Cl}_{2}\) solution c. \(0.100 \mathrm{~L}\) of a \(0.150 \mathrm{M} \mathrm{KCl}\) solution using an \(8.00 \mathrm{M} \mathrm{KCl}\) solution

Short answer

a. 12.75 mL, b. 11.92 mL, c. 1.875 mL

Step-by-step solution

Identify the dilution equation

Use the dilution equation: \[ C_1 V_1 = C_2 V_2 \]where:-\( C_1 \) is the concentration of the stock solution,-\( V_1 \) is the volume of the stock solution,-\( C_2 \) is the concentration of the diluted solution, and-\( V_2 \) is the volume of the diluted solution.

Determine known and unknown values for part (a)

Given: \( C_1 = 4.00 \) M,\( C_2 = 0.200 \) M,\( V_2 = 255 \) mLUnknown: \( V_1 \)

Solve for \( V_1 \) for part (a)

Rearrange the dilution equation to solve for \( V_1 \): \[ V_1 = \frac{C_2 V_2}{C_1} \]Substitute the given values: \[ V_1 = \frac{0.200 \times 255}{4.00} \]\[ V_1 = 12.75 \] mL

Determine known and unknown values for part (b)

Given: \( C_1 = 6.00 \) M,\( C_2 = 0.100 \) M,\( V_2 = 715 \) mLUnknown: \( V_1 \)

Solve for \( V_1 \) for part (b)

Using the same method as in Step 3, substitute the given values: \[ V_1 = \frac{0.100 \times 715}{6.00} \]\[ V_1 = 11.92 \] mL

Determine known and unknown values for part (c)

Given: \( C_1 = 8.00 \) M,\( C_2 = 0.150 \) M,\( V_2 = 100 \) mL (since 0.100 L = 100 mL)Unknown: \( V_1 \)

Solve for \( V_1 \) for part (c)

Using the same method as earlier, substitute the given values: \[ V_1 = \frac{0.150 \times 100}{8.00} \]\[ V_1 = 1.875 \] mL

Key concepts

molarity

Molarity is a measure of the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. The formula for molarity is given by:
\[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \]
For example, if you have 1 mole of sodium chloride (NaCl) dissolved in 1 liter of water, the molarity of the NaCl solution is 1 M.
Molarity is commonly used in chemistry to describe the concentration of a solution because it allows chemists to relate the amount of solute to the volume of the solution, which is essential for reactions and dilutions.
When working with molarity, it's crucial to remember that:

• The volume of the solution should be in liters.
• The solute should be measured in moles.

This ensures accurate calculations and consistency in laboratory experiments.

stock solution

A stock solution is a highly concentrated solution from which diluted solutions are prepared. Stock solutions are usually prepared for convenience, as they can be diluted to achieve various concentrations required for different experiments.
The key characteristics of a stock solution include:

• High concentration of solute: Stock solutions have a high molarity, which means they contain a large amount of solute in a small volume of solvent.
• Stability: Stock solutions are typically chemically stable and can be stored for extended periods without significant degradation.

An important concept when using stock solutions is the dilution equation: \[ C_1 V_1 = C_2 V_2 \]
where:

• \( C_1 \) = concentration of the stock solution
• \( V_1 \) = volume of the stock solution
• \( C_2 \) = concentration of the diluted solution
• \( V_2 \) = volume of the diluted solution

This equation helps in determining the volume of stock solution needed to achieve a desired concentration when diluted. Simply plug in the known values to solve for the unknown variable.

diluted solution

A diluted solution is a solution obtained by adding more solvent to a given volume of stock solution, thereby decreasing its concentration. The process of dilution is essential in many chemical experiments to achieve the desired concentration for reactions or analyses.
Key points about diluted solutions include:

• Lower concentration: The concentration of the diluted solution is always lower than that of the stock solution.
• Increased volume: The volume of the diluted solution is greater than that of the stock solution.
• Conservation of solute: The total amount of solute remains the same before and after dilution.

Using the dilution equation:

\[ C_1 V_1 = C_2 V_2 \]
we can determine the volume of stock solution needed to prepare a specific volume of diluted solution at the desired concentration.
For example, in part (a) of the given exercise, the goal was to prepare 255 mL of a 0.200 M \( \mathrm{HNO}_3 \) solution using a 4.00 M \( \mathrm{HNO}_3 \) stock solution. By rearranging and solving the dilution equation, we found that 12.75 mL of the stock solution is required. This same method can be applied to various other dilution problems.
Diluting solutions not only help achieve the desired concentration but also ensures safety and precision in chemical experiments.