Problem 80 - General Chemistry

Question

A chemist wants to prepare \(0.50 M\) HCl. Commercial hydrochloric acid is \(12.4 M\). How many milliliters of the commercial acid does the chemist require to make up \(1.50 \mathrm{~L}\) of the dilute acid?

Step-by-Step Solution

Answer

The chemist requires 60.5 mL of commercial HCl.

Step 1: Understand the Problem

We need to dilute a concentrated solution of hydrochloric acid (HCl) with a molarity of \(12.4 \text{ M}\) to make a \(0.50 \text{ M}\) solution. The total volume of the diluted solution should be \(1.50 \text{ L}\). Our task is to calculate the volume of the concentrated solution needed.

Step 2: Apply the Dilution Formula

To find the required volume of concentrated HCl, we use the dilution formula: \(C_1V_1 = C_2V_2\), where \(C_1\) and \(V_1\) are the concentration and volume of the concentrated solution, respectively, and \(C_2\) and \(V_2\) are those of the diluted solution. Plug in the values: \(12.4V_1 = 0.50 \times 1.50\).

Step 3: Solve for Initial Volume

Rearrange the equation from Step 2 to solve for \(V_1\): \[ V_1 = \frac{0.50 \times 1.50}{12.4} \]. Calculate the right side to find \(V_1\).

Step 4: Perform the Calculation

Calculate \(V_1\) by substituting the values into the equation: \[ V_1 = \frac{0.50 \times 1.50}{12.4} \approx 0.0605 \text{ L} \]. Convert to milliliters: \(0.0605 \text{ L} = 60.5 \text{ mL}\).

Step 5: Conclusion

The chemist needs \(60.5 \text{ mL}\) of the commercial \(12.4 \text{ M}\) HCl to prepare \(1.50 \text{ L}\) of a \(0.50 \text{ M}\) HCl solution.

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It is calculated by taking the number of moles of a solute and dividing it by the volume of the solution in liters. This unit of concentration is denoted by a capital "M" and is used to describe how much of a substance is present in a given volume. For example, a solution with a molarity of \(12.4 \, M\) means there are \(12.4\) moles of solute per liter of solution. Molarity is a crucial concept in chemistry as it allows for easy calculations when diluting or mixing solutions, as seen in the problem with hydrochloric acid (HCl).

Knowing the molarity of a solution helps to ensure that reactions proceed with the correct ratios of chemicals, which is essential for accuracy in laboratory settings.

• The formula for molarity: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \).
• Useful for calculating how to prepare solutions of a desired concentration by dilution or mixing.
• Essential in predicting the outcomes of reactions since it helps determine the quantities of reactants and products.

Concentration Calculations

Concentration calculations are essential when working with solutions in chemistry. It guides you in determining how much of one substance is mixed with a specific volume of another substance, usually a liquid. These calculations are vital when performing dilutions, where you decrease the concentration of a solution by adding more solvent to it.

In the dilution process, a key formula is used: \( C_1V_1 = C_2V_2 \), where:

• \( C_1 \) is the concentration of the concentrated solution.
• \( V_1 \) is the volume of the concentrated solution.
• \( C_2 \) is the concentration of the dilute solution.
• \( V_2 \) is the volume of the dilute solution.

By rearranging this formula, you can solve for any unknown (in this case, \( V_1 \)), to find out how much of the concentrated solution you need to make a new solution of the desired concentration. This method ensures that the total number of moles of solute remains constant before and after dilution.

These calculations are particularly useful in laboratory settings, where precise concentrations are needed for experiments and reactions.

Solution Preparation

Solution preparation is a common task in chemistry that involves making a specific concentration of a solution from another, often more concentrated, solution. The process requires careful planning to ensure the correct ratios are used, maintaining the integrity and intended strength of the solution.

When preparing a solution, it often involves dilution, which is adding a solvent to a concentrated solution to decrease its molarity to a desired level. Using the example of hydrochloric acid (HCl), preparation involved calculating the volume of the 12.4 M concentrated acid needed to make a 1.50 L of 0.50 M solution by using the dilution formula: \( C_1V_1 = C_2V_2 \).

Important steps in solution preparation include:

• Determining the desired concentration and volume of the final solution.
• Calculating the amount of the concentrated solution required using concentration equations.
• Mixing carefully to ensure homogeneity, often adding the solvent gradually to control the dilution process.

Solution preparation ensures experiments are performed consistently and results are reliable. It's a foundation for many chemical reactions, assays, and tests conducted in various scientific fields.