Problem 73 - Chemistry The Central Science In Si Units

Question

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{MNH}_{3} ?\) (b) If you take a 10.0 -mL portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

Step-by-Step Solution

Answer

(a) Approximately 16.89 mL of the stock solution is needed. (b) The concentration of the new solution is 0.296 M.

Step 1: Understand the Problem for Part (a)

We need to determine the volume of a 14.8 M NH₃ solution required to make 1000.0 mL of a 0.250 M NH₃ solution. This can be addressed using the dilution formula.

Step 2: Apply the Dilution Formula for Part (a)

Use the formula: \[C_1V_1 = C_2V_2\]where \(C_1 = 14.8\, \mathrm{M}\), \(C_2 = 0.250\, \mathrm{M}\), and \(V_2 = 1000.0\, \mathrm{mL}\). Solve for \(V_1\) to find the volume of the stock solution needed.

Step 3: Calculate the Volume for Part (a)

Re-arrange the formula: \[V_1 = \frac{C_2V_2}{C_1}\]Substitute the values:\[V_1 = \frac{0.250 \times 1000.0}{14.8}\, \mathrm{mL} = \frac{250.0}{14.8}\, \mathrm{mL} \approx 16.89\, \mathrm{mL}\]Thus, you need approximately 16.89 mL of the 14.8 M NH₃ stock solution.

Step 4: Understand the Problem for Part (b)

We need to find the concentration of the NH₃ solution after diluting 10.0 mL of the stock solution to a total volume of 0.500 L.

Step 5: Apply the Dilution Formula for Part (b)

Use the same dilution formula: \[C_1V_1 = C_2V_2\]where \(C_1 = 14.8\, \mathrm{M}\), \(V_1 = 10.0\, \mathrm{mL}\), and \(V_2 = 0.500 \times 1000 = 500\, \mathrm{mL}\). Solve for \(C_2\) to find the new concentration.

Step 6: Calculate the Concentration for Part (b)

Re-arrange the formula: \[C_2 = \frac{C_1V_1}{V_2}\]Substitute the values:\[C_2 = \frac{14.8 \times 10.0}{500}\, \mathrm{M} = \frac{148.0}{500}\, \mathrm{M} = 0.296\, \mathrm{M}\]The concentration of the diluted solution is 0.296 M NH₃.

Key Concepts

Molarity Calculation

To fully understand how to calculate molarity, it's important to know it describes the concentration of a solution. Molarity, denoted as \( M \), is defined as the number of moles of solute per liter of solution. This means it measures the amount of a substance within a given volume. Calculating molarity involves a straightforward formula:

• \( M = \frac{n}{V} \)

Here, \( n \) is the number of moles of the solute and \( V \) is the volume of the solution in liters. When preparing or analyzing solutions, knowing the molarity is essential because it allows you to understand the strength of the solution. By using the molarity formula, you can convert between the mass of solute and its volume within a solution. This conversion is particularly useful in different chemical calculations, such as those involving dilution formulas.

Solution Concentration

Solution concentration refers to how much solute is present in a given amount of solvent or solution. It's a crucial aspect in chemistry that determines how a solution will react in processes or experiments. Concentration is often expressed in terms of molarity, which makes it easy to calculate and adjust in laboratory settings. Different ways to express concentration include:

• Percent concentration: the amount of solute divided by the total solution amount, multiplied by 100.
• Molality: moles of solute per kilogram of solvent.
• Mole fraction: the ratio of moles of one component to the total moles of all components.

The concentration directly affects the behavior of molecules in a solution, their reactivity, and properties like boiling and freezing points. It is crucial to measure and adjust concentrations to ensure that reactions occur as expected.

Volume Conversion

In preparing or analyzing solutions, converting volumes from one unit to another is vital. This skill helps you accurately measure how much solvent or solute is needed to achieve a desired concentration. Volume can be expressed in various units such as liters (L), milliliters (mL), or microliters (µL). Here's a common conversion method:

• 1 liter (L) = 1000 milliliters (mL)
• 1 milliliter (mL) = 1000 microliters (µL)

Being comfortable with these conversions allows you to correctly relate the amount of solute needed for solutions of various volumes. For example, if you're tasked with converting 0.500 L into milliliters for a dilution, you recall that 0.500 L equals 500 mL. Accurate volume conversion is essential for applying dilution formulas correctly, ensuring the preparation of solutions with precise concentrations.