Question

Calculate the concentration of each of the following diluted solutions: a. \(2.0 \mathrm{~L}\) of a \(6.0 \mathrm{M} \mathrm{HCl}\) solution is added to water so that the final volume is \(6.0 \mathrm{~L}\). b. Water is added to \(0.50 \mathrm{~L}\) of a \(12 \mathrm{M} \mathrm{NaOH}\) solution to make \(3.0 \mathrm{~L}\) of a diluted \(\mathrm{NaOH}\) solution. c. A \(10.0-\mathrm{mL}\) sample of a \(25 \%(\mathrm{~m} / \mathrm{v})\) KOH solution is diluted with water so that the final volume is \(100.0 \mathrm{~mL}\). d. A \(50.0\) -mL sample of a \(15 \%(\mathrm{~m} / \mathrm{v}) \mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to water to give a final volume of \(250 \mathrm{~mL}\).

Short answer

a. 2.0 M b. 2.0 M c. 2.5 % (m/v) d. 3.0 % (m/v)

Step-by-step solution

Identify the dilution formula

The dilution formula is given by: \ \ \ \ \[\text{Concentration}_{1} \times \text{Volume}_{1} = \text{Concentration}_{2} \times \text{Volume}_{2}\ \] This formula can be used to find the concentration of the diluted solution given the initial concentration and volumes before and after dilution.

Calculate the concentration (part a)

Given \(\text{Concentration}_{1} = 6.0 \mathrm{M}\) and \( \text{Volume}_{1} = 2.0 \mathrm{~L}\), final volume \( \text{Volume}_{2} = 6.0 \mathrm{~L}\)\Set up the equation and solve for \(\text{Concentration}_{2}\): \[\text{Concentration}_{1} \times \text{Volume}_{1} = \text{Concentration}_{2} \times \text{Volume}_{2}\ \Rightarrow 6.0 \mathrm{M} \times 2.0 \mathrm{L} = \text{Concentration}_{2} \times 6.0 \mathrm{L} \] \ \Therefore: \[\text{Concentration}_{2} = \frac{6.0 \mathrm{M} \times 2.0 \mathrm{L}}{6.0 \mathrm{L}} = 2.0 \mathrm{M} \]

Calculate the concentration (part b)

Given \(\text{Concentration}_{1} = 12 \mathrm{M}\) and \( \text{Volume}_{1} = 0.50 \mathrm{~L}\), final volume \( \text{Volume}_{2} = 3.0 \mathrm{~L}\)\Set up the equation and solve for \(\text{Concentration}_{2}\): \[\text{Concentration}_{1} \times \text{Volume}_{1} = \text{Concentration}_{2} \times \text{Volume}_{2}\ \Rightarrow 12 \mathrm{M} \times 0.50 \mathrm{L} = \text{Concentration}_{2} \times 3.0 \mathrm{L} \] \ \Therefore: \[\text{Concentration}_{2} = \frac{12 \mathrm{M} \times 0.50 \mathrm{L}}{3.0 \mathrm{L}} = 2.0 \mathrm{M} \]

Calculate the concentration (part c)

Given \( \text{Concentration}_{1} = 25 \%(\text{m}/\text{v}) \) and \( \text{Volume}_{1} = 10.0 \mathrm{~mL}\), final volume \( \text{Volume}_{2} = 100.0 \mathrm{~mL}\)\Set up the equation and solve for \(\text{Concentration}_{2}\): \[\text{Concentration}_{1} \times \text{Volume}_{1} = \text{Concentration}_{2} \times \text{Volume}_{2}\ \Rightarrow 25 \%(\text{m}/\text{v}) \times 10.0 \mathrm{mL} = \text{Concentration}_{2} \times 100.0 \mathrm{mL} \] \ \Therefore: \[\text{Concentration}_{2} = \frac{25 \%(\text{m}/\text{v}) \times 10.0 \mathrm{mL}}{100.0 \mathrm{mL}} = 2.5 \%(\text{m}/\text{v}) \]

Calculate the concentration (part d)

Given \( \text{Concentration}_{1} = 15 \%(\text{m}/\text{v}) \) and \( \text{Volume}_{1} = 50.0 \mathrm{~mL}\), final volume \( \text{Volume}_{2} = 250 \mathrm{~mL}\)\Set up the equation and solve for \(\text{Concentration}_{2}\): \[\text{Concentration}_{1} \times \text{Volume}_{1} = \text{Concentration}_{2} \times \text{Volume}_{2}\ \Rightarrow 15 \%(\text{m}/\text{v}) \times 50.0 \mathrm{mL} = \text{Concentration}_{2} \times 250 \mathrm{mL} \] \ \Therefore: \[\text{Concentration}_{2} = \frac{15 \%(\text{m}/\text{v}) \times 50.0 \mathrm{mL}}{250 \mathrm{mL}} = 3.0 \%(\text{m}/\text{v}) \]

Key concepts

Molarity

Molarity is a fundamental concept in chemistry that measures the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution (mol/L). The formula to calculate molarity is:
\[M = \frac{n}{V}\]
where \(M\) is molarity, \(n\) is the number of moles of solute, and \(V\) is the volume of solution in liters. This measure helps in understanding how concentrated a chemical solution is, which is critical for reaction rates and equilibrium calculations.

Concentration

Concentration refers to the amount of solute dissolved in a given quantity of solvent or solution. It can be expressed in various ways, such as molarity (M), mass/volume percentage (m/v %), or parts per million (ppm). Knowing the concentration of solutions is important in fields like pharmacology, environmental science, and chemical engineering because it affects the properties and reactions of the solution. For instance, in the exercise, we calculate the concentration of diluted solutions after water is added.

Dilution Formula

The dilution formula is a simple equation used to calculate the concentration of a solution after it has been diluted. It states that the product of the initial concentration and volume is equal to the product of the final concentration and volume:
\[C_1 V_1 = C_2 V_2\]
Here, \(C_1\) is the initial concentration, \(V_1\) is the initial volume, \(C_2\) is the final concentration, and \(V_2\) is the final volume. This formula assumes that the amount of solute remains constant during the dilution process and is a handy tool when working with solutions in laboratory settings.

Chemical Solutions

A chemical solution is a homogeneous mixture composed of two or more substances. In a solution, the solute is the substance that is dissolved, and the solvent is the substance in which the solute dissolves. For example, in a saltwater solution, salt is the solute, and water is the solvent. The properties of solutions, such as boiling point, freezing point, and vapor pressure, depend on the concentration of the solute. Understanding these properties is crucial for various applications, including chemical reactions and industrial processes.

Volume Calculations

Volume calculations are essential in working with solutions, particularly when preparing and diluting solutions. Accurately measuring volumes ensures the correct concentration of the solution. Common units of volume include liters (L), milliliters (mL), and cubic meters (m³). To convert between these units, remember that:

• 1 L = 1000 mL
• 1 m³ = 1000 L

Volume calculations are fundamental in chemistry for tasks like diluting concentrated solutions and determining the amount of solvent needed to achieve a desired concentration, as demonstrated in the given exercise.