Problem 73 - Chemistry The Central Science

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 \(\mathrm{MNH}_{3} ?(\mathbf{b})\) 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

To prepare 1000.0 mL of 0.250 M NH3 solution, you should dilute 16.9 mL of the 14.8 M NH3 stock solution. After diluting a 10.0 mL portion of the stock solution to a total volume of 0.500 L, the final concentration will be 0.296 M NH3.

Step 1: Identify the given information

Here is the given information for part (a): - Stock solution concentration (M1): 14.8 M NH3 - Desired diluted solution concentration (M2): 0.250 M NH3 - Desired diluted solution volume (V2): 1000.0 mL Our goal is to find the volume of the stock solution needed to make the diluted solution (V1).

Step 2: Use the dilution formula to solve for V1

We will use the dilution formula M1V1 = M2V2 to solve for V1: \(14.8 \times V1 = 0.250 \times 1000.0\) Now we can solve for V1: \(V1 = \frac{0.250 \times 1000.0}{14.8}\)

Step 3: Calculate V1

Plug in the values and calculate V1: \(V1 = \frac{250}{14.8} \approx 16.9 \mathrm{mL}\) So, 16.9 mL of the stock solution should be diluted to make 1000.0 mL of 0.250 M NH3 solution. Part (b):

Step 1: Identify the given information for part (b)

The given information for part (b) is: - Stock solution volume (V1): 10.0 mL - Final diluted solution volume (V2): 0.500 L (we need to convert this to mL) - Stock solution concentration (M1): 14.8 M NH3 Our goal is to find the concentration of the final solution (M2).

Step 2: Convert final volume to mL

We'll convert the final solution volume (V2) from L to mL: V2 = 0.500 L x 1000 = 500 mL

Step 3: Rearrange the dilution formula to solve for M2

Use the dilution formula M1V1 = M2V2 and rearrange it to solve for M2: \(M2 = \frac{M1 \times V1}{V2}\)

Step 4: Calculate M2

Plug in the values and calculate M2: \(M2 = \frac{14.8 \times 10.0}{500} \) M2 = 0.296 M NH3 Hence, the concentration of the final solution after diluting the 10.0 mL stock solution to a total volume of 0.500 L will be 0.296 M NH3.

Key Concepts

Molarity

Molarity is an essential concept in chemistry for expressing the concentration of a solution. It's represented by the symbol 'M' and signifies moles of solute per liter of solution. Here's what you need to remember about molarity:

• Definition: Molarity is the number of moles of a solute in one liter of solution, with the formula: \[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
• Units: Since molarity includes moles and liters, it's often expressed in terms of mol/L (Molar).
• Application: It's used to determine how concentrated a solution is, which influences rates of reaction, osmotic pressure, and other properties.

This concept allows you to perform various calculations, preparing solutions of desired concentrations for different applications in labs and industries. Always remember, knowing the molarity helps in assessing the solution's strength.

Concentration Calculations

Concentration calculations involve determining the amount of solute in a given volume of solvent. It's critical for preparing precise solutions in laboratory and industrial settings. To perform these calculations:

• Identify Key Values: Start by noting down the initial concentration (\( M_1 \)), the volume of the solution to be prepared (\( V_2 \)), and the desired concentration (\( M_2 \)).
• Use the Dilution Formula: The formula \( M_1 \times V_1 = M_2 \times V_2 \) helps relate the concentrations and volumes before and after dilution.
• Rearrange as Necessary: Depending on what you need (volume or concentration), rearrange the formula to solve:
  • For finding initial volume (\( V_1 \)), use \( V_1 = \frac{M_2 \times V_2}{M_1} \).
  • For finding final concentration (\( M_2 \)), use \( M_2 = \frac{M_1 \times V_1}{V_2} \).

These calculations are the backbone of creating solutions with specific properties, ensuring consistency and desired outcomes in experiments.

Solution Preparation

Solution preparation involves creating a solution with a desired concentration by mixing a specific amount of solute with a solvent. Here’s how you can master this process:

• Understand Your Components: Know the type of solute and solvent. Solutes are what you’re dissolving. Solvents are the liquid part, usually water.


• Calculate Proper Ratios: With the help of molarity and concentration calculations, determine how much solute to add to reach the desired solution volume.


• Use Proper Technique: Always add solute to solvent gradually, not the other way around, for better mixing and accuracy in concentration.


• Safety and Accuracy: Wear protective gear and use precise measuring tools like pipettes or graduated cylinders to ensure safety and correctness.

Preparing solutions accurately is crucial in achieving precise and consistent results in chemical reactions and scientific research. Proper preparation enhances the reliability of your experiment's outcome.