Problem 74 - Chemistry: The Central Science

Question

(a) How many milliliters of a stock solution of \(6.0 \mathrm{MHNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3}\) ? (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L}\), what will be the concentration of the diluted solution?

Step-by-Step Solution

Answer

(a) To prepare \(110 mL\) of \(0.500 M HNO_3\), you would need approximately \(9.17 mL\) of the \(6.0 M HNO_3\) stock solution. (b) The concentration of the diluted solution when \(10.0 mL\) of the stock solution is diluted to a final volume of \(0.250 L\) will be \(0.24 M\).

Step 1: Use the dilution formula for part (a)

In order to find the volume of the stock solution required to prepare \(110 mL\) of \(0.500 M HNO_3\), we can use the following formula:\[M_1V_1 = M_2V_2\]For part (a), we have: - Initial molarity (\(M_1\)) of the stock solution: \(6.0 M\) - Final molarity (\(M_2\)) of the diluted solution: \(0.500 M\) - Final volume (\(V_2\)) of the diluted solution: \(110 mL\) Our task is to find the initial volume (\(V_1\)) needed:

Step 2: Solve for the initial volume for part (a)

Plugging in the values from Step 1 into the dilution formula, we get:\[6.0 M \times V_1 = 0.500 M \times 110 mL\]Divide both sides by \(6.0 M\), and we have:\[V_1 = \frac{0.500 M \times 110 mL}{6.0 M}\]Calculate the result to find the initial volume:

Step 3: Calculate the initial volume for part (a)

Perform the calculation for \(V_1\):\[V_1 = \frac{0.500 \times 110}{6.0} = 9.166... ≈ 9.17 mL\]To prepare \(110 mL\) of \(0.500 M HNO_3\), you would need approximately \(9.17 mL\) of the \(6.0 M HNO_3\) stock solution.

Step 4: Use the dilution formula again for part (b)

Now, we need to find the concentration of the solution when we dilute \(10.0 mL\) of the stock solution to a final volume of \(0.250 L\). We can use the dilution formula again: For part (b), we have: - Initial molarity (\(M_1\)) of the stock solution: \(6.0 M\) - Initial volume (\(V_1\)) of the stock solution: \(10.0 mL\) - Final volume (\(V_2\)) of the diluted solution: \(0.250 L = 250 mL\) Our task is to find the final molarity (\(M_2\)) of the diluted solution:

Step 5: Solve for the final molarity for part (b)

Plugging in the values from Step 4 into the dilution formula, we get:\[6.0 M \times 10.0 mL = M_2 \times 250 mL\]Divide both sides by \(250 mL\), and we have:\[M_2 = \frac{6.0 M \times 10.0 mL}{250 mL}\]Calculate the final molarity of the diluted solution:

Step 6: Calculate the final molarity for part (b)

Perform the calculation for \(M_2\):\[M_2 = \frac{6.0 \times 10.0}{250} = 0.24 M\]So, the concentration of the diluted solution will be \(0.24 M\).

Key Concepts

Molarity Calculation

Molarity is one of the most fundamental concepts in chemistry when dealing with solutions. It represents how concentrated a solution is by measuring the number of moles of solute present in one liter of solution. Molarity is denoted by the symbol \( M \) and is calculated using the equation:

• \( M = \frac{n}{V} \), where \( n \) is the number of moles of solute and \( V \) is the volume of solution in liters.

When working with problems of solution concentration, it is essential to convert any given volumes into liters if they're not already, ensuring that units match throughout the calculations.
For example, in a dilution problem, if you have a stock solution with a known molarity, you can use this formula to find out how much of it is needed to make a solution with a different molarity. Understanding this concept helps in the preparation of a solution and its dilution to required concentrations.

Stock Solution

A stock solution is a concentrated form of a chemical solution that is diluted to lower concentrations for actual use. These solutions are often prepared to have high molarity so that they can be diluted multiple times for different experiments or applications.
Using stock solutions is efficient because:

• They save storage space since a small volume of a concentrated solution can make a large amount of more dilute solutions.
• Stock solutions reduce preparation time for commonly used reagents.
• They help maintain consistency across experiments.

When performing calculations related to dilutions from stock solutions, it's crucial to know the initial concentration (molarity) of your stock solution. You will apply this information in formulas that determine how much you need to dilute to achieve your desired concentrations.

Dilution Formula

The dilution formula is a straightforward yet powerful tool used to calculate the necessary volumes or concentrations before and after mixing solutions. It is expressed as:

• \( M_1V_1 = M_2V_2 \)

where:

• \( M_1 \) is the molarity of the initial stock solution,
• \( V_1 \) is the volume of the stock solution you will use,
• \( M_2 \) is the desired molarity after dilution,
• \( V_2 \) is the final volume after dilution.

Using the dilution formula allows you to calculate any of these unknowns if the other three are provided. For example, it helps determine the volume of a stock solution needed to achieve a certain concentration after dilution, or it can be used to find the concentration of a solution after a certain amount of dilution. Understanding and using the dilution formula ensures accuracy in achieving the desired concentration in laboratory processes.