Problem 75 - Chemistry Matter and Change

Question

Stock solutions of HCl with various molarities are frequently prepared. Complete Table 14.7 by calculating the volume of concentrated, or 12\(M\) , hydrochloric acid that should be used to make 1.0 \(\mathrm{L}\) of HCl solution with each molarity listed.

Step-by-Step Solution

Answer

Use the dilution formula: \(V_1 = \frac{M_2 \times V_2}{M_1}\) to calculate the volume for each desired molarity, using \(M_1 = 12\,M\) and \(V_2 = 1\, L\).

Step 1: Understanding the Problem

We are asked to find the volume of a 12 M HCl stock solution required to prepare 1.0 L of HCl solutions with different molarities. This involves the use of the dilution formula.

Step 2: Identifying the Formula

The formula we need is the dilution equation: \[ M_1 \times V_1 = M_2 \times V_2 \]where \(M_1\) and \(V_1\) are the molarity and volume of the stock solution, and \(M_2\) and \(V_2\) are the molarity and volume of the diluted solution.

Step 3: Applying the Formula

We need to rearrange the formula to solve for \(V_1\), the volume of the stock solution:\[ V_1 = \frac{M_2 \times V_2}{M_1} \]Here, \(M_1 = 12\,M\), and \(V_2 = 1.0\,L\). We'll compute \(V_1\) for each of the molarities listed in Table 14.7.

Step 4: Calculating for Different Molarities

Given molarities are not specified, let's assume we have to calculate for several common values (e.g., 1 M, 2 M, etc.):1. **For 1 M HCl Solution:** \[ V_1 = \frac{1 \times 1.0}{12} = 0.0833\, L \] or 83.3 mL.2. **For 2 M HCl Solution:** \[ V_1 = \frac{2 \times 1.0}{12} = 0.1667\, L \] or 166.7 mL.Continue this pattern for any other required molarities.

Step 5: Finalizing the Table

Enter these calculated values of \(V_1\) into Table 14.7 corresponding to each given desired molarity for the diluted solution.

Key Concepts

Molarity

Molarity is a term that describes the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. To understand molarity deeply:

• The formula for molarity is given by: \ \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
• This concept is crucial in chemistry as it allows scientists, students, and professionals to know how concentrated a solution is.
• In practical terms, a high molarity indicates a strong concentration, which often implies greater chemical reaction potential.

In the context of the exercise, molarity is essential as it is used to determine the volume needed from a stock solution to prepare desired concentrations through dilution. It is also significant in ensuring safety and efficacy in laboratory practices.

Volume Calculation

Volume calculation is an important aspect of preparing chemical solutions from stock solutions. It involves determining the amount of a solution needed to achieve a desired concentration after dilution. Key Steps in Volume Calculation:

• The dilution formula, \ \[ M_1 \times V_1 = M_2 \times V_2 \] is crucial here, as it allows you to calculate the necessary volumes based on molarity.
• Rearrange this equation to solve for the unknown volume when diluting: \ \[ V_1 = \frac{M_2 \times V_2}{M_1} \]
• Here, \(V_1\) represents the volume of the stock solution needed, while \(V_2\) is the volume of the final solution.

Using this method, you ensure that the desired concentration is achieved accurately. Volume calculation is practical and necessary to make solutions that precisely meet experimental requirements or customer specifications.

Stock Solutions

Stock solutions are concentrated solutions used to save time and space in laboratories. By diluting a stock solution, a desired solution with lower molarity can be readily prepared. Understanding Stock Solutions:

• Stock solutions are useful for preparing a variety of concentrations using the same concentrated source, reducing waste and preparation time.
• They are commonly labeled with their exact molarity, which is crucial for calculations involving dilutions.
• Safety is a key factor when handling stock solutions, as they usually contain high concentrations of active substances.

In the exercise example, a 12 M HCl stock solution means it has 12 moles of HCl per liter. From this, various molarities (like 1 M or 2 M) are prepared by calculating the correct volumes using the dilution equation. Understanding stock solutions ensures accurate laboratory work and maintains consistency in results.