Question

Determine the concentrations of the following diluted solutions. (a) \(24.75 \mathrm{~mL}\) of \(0.1832 M \mathrm{HCl}\) diluted to \(250.0 \mathrm{~mL}\) (b) \(125 \mathrm{~mL}\) of \(1.187 M \mathrm{NaOH}\) diluted to \(0.500 \mathrm{~L}\) (c) \(10.00 \mathrm{~mL}\) of \(0.2010 M\) acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\), diluted to \(50.00 \mathrm{~mL}\)

Short answer

The final concentrations after dilution are: (a) 0.0183 M HCl, (b) 0.2975 M NaOH, and (c) 0.0402 M CH3CO2H.

Step-by-step solution

Using the Dilution Formula for HCl

For HCl, the initial molarity \( M_1 \) is 0.1832 M, the initial volume \( V_1 \) is 24.75 mL, and the final volume \( V_2 \) is 250.0 mL. Substituting these values into the dilution formula, we obtain to solve for \( M_2 \): \( M_2 = M_1V_1 / V_2 = (0.1832 M)(24.75 mL) / (250.0 mL) \).

Using the Dilution Formula for NaOH

Similarly, for NaOH, \( M_1 \) is 1.187 M, \( V_1 \) is 125 mL, and \( V_2 \) is 500 mL (or 0.500 L, since 1 L = 1000 mL). Therefore, \( M_2 = M_1V_1 / V_2 = (1.187 M)(125 mL) / (500 mL) \).

Using the Dilution Formula for CH3CO2H

Lastly, for CH3CO2H, \( M_1 \) is 0.2010 M, \( V_1 \) is 10.00 mL, and \( V_2 \) is 50.00 mL. Thus, \( M_2 = M_1V_1 / V_2 = (0.2010 M)(10.00 mL) / (50.00 mL) \).

Key concepts

Molarity

Molarity is a crucial concept in chemistry when dealing with solutions. It measures the "concentration" of a solution in terms of moles of solute per liter of solution. Molarity is often denoted by the letter "M" and is used in many calculations, including dilution problems.Molarity can be expressed mathematically as:\[M = \frac{n}{V}\]Where:

• \( M \) is the molarity in moles per liter (mol/L).
• \( n \) is the number of moles of the solute.
• \( V \) is the volume of the solution in liters.

Understanding molarity helps in determining how much solute is present in a given volume of solution, which is particularly useful in experiments requiring precise measurements. Knowing the molarity allows chemists to predict how a substance will react under various conditions.

Concentration Calculation

Concentration calculations are pivotal in chemistry to determine the amount of a solute present in a given volume of solution. These calculations become particularly important in dilution processes, where a solution is made less concentrated by adding more solvent.The formula used for dilution calculations is represented as:\[M_1 \times V_1 = M_2 \times V_2\]Where:

• \( M_1 \) is the initial molarity.
• \( V_1 \) is the initial volume.
• \( M_2 \) is the final molarity after dilution.
• \( V_2 \) is the final volume after dilution.

This equation helps students and professionals understand how concentrations change when the volume of a solution increases, reflecting the inverse relationship between them. By re-arranging this equation, you can solve for any unknown variable when given the other three variables, allowing precise computation of the "concentration".

Solution Preparation

Solution preparation is a vital skill in laboratories, involving converting a solid or concentrated liquid into a solution of desired concentration. This process relies on understanding the concept of molarity and mastering concentration calculations.Here are the basic steps involved in preparing a solution:

• Determine the desired molarity and volume of the solution.
• Calculate the amount of solute needed using the formula \( n = M \times V \), where \( n \) is the number of moles, \( M \) is the desired molarity, and \( V \) is the solution volume in liters.
• Measure and mix the solute into a suitable container.
• Add solvent slowly until the total volume reaches the intended level.
• Mix the solution thoroughly to ensure even distribution of the solute.

Practicing solution preparation not only solidifies your understanding of concentration concepts but also builds the technical skills necessary for many scientific procedures. Remaining precise and adhering to these steps ensures accuracy and consistency in experiments that depend on exact "solution preparation".