Question

Calculate the molarity of each aqueous solution: (a) \(78.0 \mathrm{~mL}\) of \(0.240 \mathrm{M} \mathrm{NaOH}\) diluted to \(0.250 \mathrm{~L}\) with water (b) \(38.5 \mathrm{~mL}\) of \(1.2 \mathrm{M} \mathrm{HNO}_{3}\) diluted to \(0.130 \mathrm{~L}\) with water

Short answer

Part (a) - 0.0749 M, Part (b) - 0.355 M

Step-by-step solution

Introduction to Molarity and Dilution Formula

Molarity (M) of a solution is defined as the number of moles of solute per liter of solution. The formula for calculating the molarity after dilution is based on the relationship: \( M_1 V_1 = M_2 V_2 \), where \( M_1 \) and \( V_1 \) are the molarity and volume before dilution, and \( M_2 \) and \( V_2 \) are the molarity and volume after dilution.

Apply the Dilution Formula for Part (a)

In this part, \( M_1 = 0.240 \mathrm{M} \), \( V_1 = 78.0 \mathrm{~mL} = 0.0780 \mathrm{~L} \), and \( V_2 = 0.250 \mathrm{~L} \). Using the dilution formula: \[ 0.240 \times 0.0780 = M_2 \times 0.250 \] Solve for \( M_2 \): \[ M_2 = \frac{0.240 \times 0.0780}{0.250} = 0.07488 \mathrm{M} \]

Apply the Dilution Formula for Part (b)

For this part, \( M_1 = 1.2 \mathrm{M} \), \( V_1 = 38.5 \mathrm{~mL} = 0.0385 \mathrm{~L} \), and \( V_2 = 0.130 \mathrm{~L} \). Using the dilution formula: \[ 1.2 \times 0.0385 = M_2 \times 0.130 \] Solve for \( M_2 \): \[ M_2 = \frac{1.2 \times 0.0385}{0.130} = 0.3554 \mathrm{M} \]

Key concepts

core concept: dilution formula

Understanding the dilution formula is key to solving many chemistry problems dealing with solutions. The dilution formula, expressed as \( M_1V_1 = M_2V_2 \), helps us find out what happens to the concentration of a solution when we add more solvent. Here, \( M_1 \) is the initial molarity, and \( V_1 \) is the initial volume. After dilution, the new molarity is \( M_2 \), and the new volume is \( V_2 \). Diluting a solution means adding more solvent, typically water, which decreases the concentration but increases the volume. This formula maintains the total number of moles of solute constant. So, to apply it:

• Identify your initial concentration and volume (\( M_1 \) and \( V_1 \)).
• Determine your final volume (\( V_2 \)).
• Use the formula \( M_1V_1 = M_2V_2 \) to solve for the final concentration (\( M_2 \)).

This simple equation helps keep track of how dilutions affect solutions, ensuring accurate preparation of solutions in the lab.

core concept: solution concentration

When talking about solution concentration, it's helpful to understand it represents how much solute is dissolved in a solvent. Molarity (\( M \)) is a common way to express concentration. It is defined as the number of moles of solute per liter of solution. To calculate molarity:

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

For example, if you have 1 mole of salt (NaCl) in 1 liter of water, the molarity is 1 M. Knowing the concentration of a solution allows chemists to predict how it will behave in reactions and to create solutions with precise properties. Always remember that diluting a solution will decrease its molarity, as the volume increases while the amount of solute stays the same.

core concept: step-by-step chemical problem solving

Breaking down chemical problems step by step makes them easier to solve. Let's tackle a typical dilution problem step by step. Take the original example where we are given different volumes and concentrations before and after dilution. Here is how to proceed:
First, identify the given values. For part (a):

• Initial concentration (\( M_1 \)) = 0.240 M
• Initial volume (\( V_1 \)) = 78.0 mL = 0.0780 L
• Final volume (\( V_2 \)) = 0.250 L

Use the dilution formula \( M_1V_1 = M_2V_2 \):
Plug in the values: \[0.240 \times 0.078 = M_2 \times 0.250\text{ L} \]Solve for \( M_2 \): \[M_2 = \frac{0.240 \times 0.078}{0.250} = 0.07488 \text{ M}\]
By following these steps carefully, you ensure accuracy and build a strong understanding of the problem-solving process. Practice this method, and soon solving such problems will become second nature.