Problem 1 - Chemical Principles in the Laboratory

Question

\(7.2 \mathrm{mL}\) of \(6.0 \mathrm{M} \mathrm{NaOH}\) are diluted with water to a volume of \(400.0 \mathrm{mL}\). You are asked to find the molarity of the resulting solution. a. First find out how many moles of NaOH there are in 7.2 mL of \(6.0 \mathrm{M}\) NaOH. Use Equation \(1 .\) Note that the volume must be in liters. b. since the total number of moles of NaOH is not changed on dilution, the molarity after dilution can also be found by Equation 1, using the final volume of the solution. Calculate that molarity.

Step-by-Step Solution

Answer

The molarity of the resulting solution is 0.108 M.

Step 1: Convert Volume to Liters

The first step is to convert the initial volume of the sodium hydroxide solution from milliliters to liters. Since there are 1000 mL in 1 L, divide the volume by 1000. Thus, \(7.2\, \text{mL} = \frac{7.2}{1000}\, \text{L} = 0.0072\, \text{L}\).

Step 2: Calculate Initial Moles of NaOH

Use the molarity formula to find the number of moles of NaOH in the initial solution. Molarity (\(M\)) is defined as moles of solute per liter of solution, so: \[\text{moles of NaOH} = \text{Molarity} \times \text{Volume} \]Substitute the known values: \( \text{moles of NaOH} = 6.0\, \text{M} \times 0.0072\, \text{L} = 0.0432\, \text{mol}\).

Step 3: Calculate Final Molarity after Dilution

Since moles of solute remain constant during dilution, use the second molarity formula with the final total volume of the solution to find the new molarity: \[ \text{Molarity after dilution} = \frac{\text{moles of solute}}{\text{final volume in liters}} \]Convert the final volume to liters: \( 400.0\, \text{mL} = \frac{400.0}{1000}\, \text{L} = 0.400\, \text{L} \).Now, substitute: \[ \text{Molarity after dilution} = \frac{0.0432\, \text{mol}}{0.400\, \text{L}} = 0.108\, \text{M} \].

Key Concepts

Molarity Calculation

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. Think of it as a measure of how many moles of a substance are dissolved in a given volume of solution. To calculate molarity, we use the formula:

• \( M = \frac{n}{V} \)

where \( M \) is molarity in moles per liter (M), \( n \) is the number of moles, and \( V \) is the volume of the solution in liters.
In the initial step of our exercise, we used this formula to determine the number of moles of NaOH in the solution. This helps us understand how concentration is affected when a solution is diluted. By multiplying molarity (6.0 M) by the converted volume (0.0072 L), we find that there are 0.0432 moles of NaOH initially present. Remember, the molarity formula is key for solving problems involving concentration and dilution.

Volume Conversion

Volume conversion is crucial in chemistry as it often involves changing measurements from one unit to another. In our exercise, we needed to convert milliliters to liters. This is a typical conversion because molarity calculations are done using liters as the standard unit of volume.
To convert milliliters to liters, divide the milliliters by 1000:

• Initial conversion: \( 7.2 \, \text{mL} = \frac{7.2}{1000} \, \text{L} = 0.0072 \, \text{L} \)
• Final conversion: \( 400.0 \, \text{mL} = \frac{400.0}{1000} \, \text{L} = 0.400 \, \text{L} \)

This conversion ensures that when calculating molarity, the units are consistent, avoiding errors that could occur if different units are mixed. Accurately converting volumes allows us to correctly determine and compare the concentration before and after dilution.

Mole Calculation

Mole calculations are fundamental in chemistry. They help us relate the macroscopic world we measure to the microscopic world of atoms and molecules. In our example, calculating the moles of NaOH allowed us to establish the amount of solute in the solution, crucial in understanding how dilution affects concentration.
Using the formula for moles:

• \( \text{moles} = \text{Molarity} \times \text{Volume} \)

This formula lets us calculate how many moles of NaOH are present by using its molarity and volume.Substituting in our values gives:

• \( 0.0432 \, \text{mol} = 6.0 \, \text{M} \times 0.0072 \, \text{L} \)

Understanding these mole calculations helps us when analyzing reactions and preparing solutions, as it bridges the quantity in grams (often measured) to its chemical composition. During dilution, although the volume of the solution increases, this mole amount remains constant, which is crucial for computing the final diluted molarity accurately.