Problem 24 - Chemistry Matter and Change

Question

What volume of a 3.00\(M \mathrm{Kl}\) stock solution would you use to make 0.300 \(\mathrm{L}\) of a 1.25 \(\mathrm{M} \mathrm{Kl}\) solution?

Step-by-Step Solution

Answer

0.125 L

Step 1: Understand the Problem

We need to calculate the volume of a stock solution that has a concentration (molarity) of 3.00 M needed to make 0.300 L of a solution with a desired concentration of 1.25 M.

Step 2: Use the Dilution Formula

The formula to use for dilution problems is \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( V_1 \) are the concentration and volume of the stock solution, and \( C_2 \) and \( V_2 \) are the concentration and volume of the diluted solution.

Step 3: Substitute Known Values

Substitute the known values into the dilution formula: \( 3.00 \, ext{M} \times V_1 = 1.25 \, ext{M} \times 0.300 \, ext{L} \).

Step 4: Solve for \( V_1 \)

Solve the equation for \( V_1 \):\[V_1 = \frac{1.25 \, ext{M} \times 0.300 \, ext{L}}{3.00 \, ext{M}} \]Calculate the result: \[V_1 = \frac{0.375 \, ext{M} \cdot ext{L}}{3.00 \, ext{M}} = 0.125 \, ext{L} \]

Step 5: Conclusion

Thus, you need 0.125 L of the 3.00 M stock solution to make 0.300 L of a 1.25 M solution.

Key Concepts

Molarity

Molarity, often represented by the symbol "M," is a way to express the concentration of a solute in a solution. It's common in chemistry, and it measures the number of moles of solute present per liter of solution. For example, a 3.00 M solution means there are 3 moles of solute—here, likely potassium iodide (Kl)—dissolved in one liter of total solution. Understanding molarity is crucial because it allows chemists to precisely predict and control the outcome of reactions.
In practical terms, knowing molarity helps you understand how much of a substance is available to react. This can be particularly important in laboratory settings where specific concentration levels are necessary to achieve desired chemical reactions or to carry out experiments accurately.

• Molarity (M) = moles of solute / liters of solution
• Higher molarity means more solute in the solution.
• It's crucial for predicting reaction outcomes.

Mastering molarity is a foundational skill for solving dilution problems and understanding chemical equilibria.

Stock solution

A stock solution is a concentrated solution that can be diluted to achieve a lower concentration. It is often used to save space and to make experiments more efficient. In laboratories, stock solutions are prepared once in higher concentrations to be used in multiple experiments, thus reducing the need for repeated preparation.
An important advantage of using a stock solution is consistency. When you dilute a stock solution, you can ensure that every new solution has the same starting composition, allowing for repeatable and reliable results.
Imagine if you need multiple batches of a certain solution; preparing it from a stock solution each time ensures that all batches have uniform concentrations.

• Stock solutions save space and preparation time.
• They ensure consistency across experiments.
• Aids in conducting multiple experiments with the same reagents without having to constantly re-measure and re-dissolve the solute.

In the given problem, the 3.00 M solution is your stock solution, which you use to create a less concentrated solution.

Dilution formula

The dilution formula is a simple and powerful tool used to determine how to dilute a solution to a desired concentration. It can be represented as: \[ C_1V_1 = C_2V_2 \]Here,

• \(C_1\) is the concentration (molarity) of the stock solution.
• \(V_1\) is the volume of the stock solution needed.
• \(C_2\) is the concentration of the final diluted solution.
• \(V_2\) is the volume of the final diluted solution.

Using this formula, you can solve many chemical problems related to titration and preparation of solutions. It allows you to scale solutions up or down predictably, meaning you can tailor the concentration to the needs of a specific procedure or experiment.
In the problem, by substituting the known values: \( C_1 = 3.00 \, \text{M} \), \( C_2 = 1.25 \, \text{M} \), and \( V_2 = 0.300 \, \text{L} \), you can calculate \( V_1 \), the volume of the stock solution required, as:\[ V_1 = \frac{1.25 \, \text{M} \times 0.300 \, \text{L}}{3.00 \, \text{M}} = 0.125 \, \text{L} \] This means you need 0.125 L of the 3.00 M stock solution to create the needed 1.25 M solution.