Problem 29 - Chemistry The Science in Context

Question

Calculate the final concentrations of the following aqueous solutions after each has been diluted to a final volume of \(25.0 \mathrm{mL}:\) a. \(3.00 \mathrm{mL}\) of \(0.175 M \mathrm{K}^{+}\) b. \(2.50 \mathrm{mL}\) of \(10.6 \mathrm{m} M \mathrm{LiCl}\) c. \(15.00 \mathrm{mL}\) of \(7.24 \times 10^{-2} \mathrm{m} M \mathrm{Zn}^{2+}\)

Step-by-Step Solution

Answer

Question: Calculate the final concentrations of the following solutions after dilution: a) A 3.00 mL solution of 0.175 M K+ is diluted to 25.0 mL. b) A 2.50 mL solution of 10.6 mM LiCl is diluted to 25.0 mL. c) A 15.00 mL solution of 7.24 x 10^-2 M Zn2+ is diluted to 25.0 mL. Answer: a) The final concentration of the K+ solution is 0.021 M. b) The final concentration of the LiCl solution is 0.00106 M. c) The final concentration of the Zn2+ solution is 4.34 x 10^-2 M.

Step 1: Identify the given values

The initial volume of the solution, \(V_1 = 3.00 \mathrm{mL}\), the initial molarity, \(M_1 = 0.175 M\), and the final volume, \(V_2 = 25.0 \mathrm{mL}\).

Step 2: Apply the dilution formula

Using the formula \(M_1V_1 = M_2V_2\), let's plug in the values: \((0.175 M)(3.00 \mathrm{mL}) = M_2(25.0 \mathrm{mL})\).

Step 3: Solve for the final molarity

Solve for \(M_2\): \(M_2 = \frac{(0.175 M)(3.00 \mathrm{mL})}{25.0 \mathrm{mL}} = 0.021 M\). The final concentration of the \(\mathrm{K}^+\) solution is \(0.021 M\). b.

Step 1: Identify the given values

The initial volume of the solution, \(V_1 = 2.50 \mathrm{mL}\), the initial molarity, \(M_1 = 0.0106 M\) (note: converted from \(10.6 \mathrm{m} M\) to \(\mathrm{M}\) by multiplying by \(10^{-3}\)), and the final volume, \(V_2 = 25.0 \mathrm{mL}\).

Step 2: Apply the dilution formula

Using the formula \(M_1V_1 = M_2V_2\), let's plug in the values: \((0.0106 M)(2.50 \mathrm{mL}) = M_2(25.0 \mathrm{mL})\).

Step 3: Solve for the final molarity

Solve for \(M_2\): \(M_2 = \frac{(0.0106 M)(2.50 \mathrm{mL})}{25.0 \mathrm{mL}} = 0.00106 M\). The final concentration of the \(\mathrm{LiCl}\) solution is \(0.00106 M\). c.

Step 1: Identify the given values

The initial volume of the solution, \(V_1 = 15.00 \mathrm{mL}\), the initial molarity, \(M_1 = 7.24 \times 10^{-2} M\), and the final volume, \(V_2 = 25.0 \mathrm{mL}\).

Step 2: Apply the dilution formula

Using the formula \(M_1V_1 = M_2V_2\), let's plug in the values: \((7.24 \times 10^{-2} M)(15.00 \mathrm{mL}) = M_2(25.0 \mathrm{mL})\).

Step 3: Solve for the final molarity

Solve for \(M_2\): \(M_2 = \frac{(7.24 \times 10^{-2} M)(15.00 \mathrm{mL})}{25.0 \mathrm{mL}} = 4.34 \times 10^{-2} M\). The final concentration of the \(\mathrm{Zn}^{2+}\) solution is \(4.34 \times 10^{-2} M\).

Key Concepts

Understanding Molarity

Molarity is one of the most important concepts in chemistry. It is a measure of the concentration of a solute in a solution. Specifically, molarity is defined as the number of moles of solute divided by the liters of solution. This is represented by the formula:\[M = \frac{n}{V}\]where:

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

It is crucial to note that molarity tells us how much of a substance is present in a given volume of solution. It is not just about the amount of solute but also about how it is distributed. This concept is vital when performing dilution calculations, as it helps to predict how a solution will react or behave in different scenarios.For example, if you have a 2 M solution of NaCl, it means you have 2 moles of NaCl in every liter of solution. This concentration can change with the addition of more solvent, which is where dilution comes into play.

Exploring Aqueous Solutions

An aqueous solution is any solution where water acts as the solvent. The word 'aqueous' comes from 'aqua', which means water. In chemistry, aqueous solutions are incredibly common and important because water is such a versatile solvent. It can dissolve a wide range of substances, making it an ideal medium for chemical reactions. Water’s polarity and ability to form hydrogen bonds allow it to dissolve ionic compounds and polar molecules effectively. When a substance dissolves in water, it dissociates into ions or molecules, dispersing uniformly throughout the solvent. This creates a homogeneous mixture known as an aqueous solution. In everyday life, common aqueous solutions include things like saltwater, which contains dissolved NaCl, and sugar in water. In laboratory settings, many chemical reactions are conducted in aqueous solutions due to water's ability to facilitate the interactions between different compounds.

The Dilution Formula

Dilution is the process of decreasing the concentration of a solute in a solution, usually by adding more solvent. The dilution formula is a handy tool to calculate the concentration of a solution after it has been diluted:\[M_1V_1 = M_2V_2\]In this formula:

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

This equation works because the amount of solute remains constant before and after dilution; only the volume of the solvent changes. Therefore, the product of the initial molarity and volume equals the product of the final molarity and volume.Using the dilution formula helps you determine how much solvent you need to add to achieve a desired concentration in laboratory or industrial settings. For example, if you have a highly concentrated acid and you need a specific weaker concentration, you can calculate exactly how much water to add to obtain that concentration, consistently ensuring safe and accurate results.