Problem 29 - Chemistry: An Atoms-Focused Approach

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. \(1.00 \mathrm{mL}\) of \(0.452 M \mathrm{Na}^{+}\) b. \(2.00 \mathrm{mL}\) of \(3.4 \mathrm{m} M \mathrm{LiCl}\) c. \(5.00 \mathrm{mL}\) of \(6.42 \times 10^{-2} \mathrm{m} M \mathrm{Zn}^{2+}\)

Step-by-Step Solution

Answer

a. Na+ ions with an initial concentration of 0.452 M and an initial volume of 1.00 mL. b. LiCl with an initial concentration of 3.40 mM and an initial volume of 2.00 mL. c. Zn2+ ions with an initial concentration of 6.42 x 10^-2 mM and an initial volume of 5.00 mL. Answer: a. The final concentration of Na+ ions after dilution is 0.01808 M. b. The final concentration of LiCl after dilution is 2.72 x 10^-4 M. c. The final concentration of Zn2+ ions after dilution is 1.284 x 10^-2 mM.

Step 1: a. Final concentration of Na+ ions

In this case, we are given the initial concentration (C1) as 0.452 M and initial volume (V1) as 1.00 mL, and we need to find the final concentration (C2) after dilution to a final volume (V2) of 25.0 mL. We can use the dilution formula for this: \(C1 \times V1 = C2 \times V2\) \((0.452 \mathrm{M}) \times (1.00 \mathrm{mL}) = C2 \times (25.0 \mathrm{mL})\) Now, we can solve for C2: \(C2 = \dfrac{(0.452 \mathrm{M}) \times (1.00 \mathrm{mL})}{(25.0 \mathrm{mL})}\) \(C2 = 0.01808 \mathrm{M}\) Thus, the final concentration of Na+ ions after dilution is 0.01808 M.

Step 2: b. Final concentration of LiCl

In this case, we are given the initial concentration (C1) as 3.40 mM and initial volume (V1) as 2.00 mL. We need to find the final concentration (C2) after dilution to a final volume (V2) of 25.0 mL. Again, we can use the dilution formula for this: \(C1 \times V1 = C2 \times V2\) \((3.4 \times 10^{-3} \mathrm{M}) \times (2.00 \mathrm{mL}) = C2 \times (25.0 \mathrm{mL})\) Now, we can solve for C2: \(C2 = \dfrac{(3.4 \times 10^{-3} \mathrm{M}) \times (2.00 \mathrm{mL})}{(25.0 \mathrm{mL})}\) \(C2 = 2.72 \times 10^{-4} \mathrm{M}\) Thus, the final concentration of LiCl after dilution is \(2.72 \times 10^{-4} \mathrm{M}\).

Step 3: c. Final concentration of Zn2+ ions

In this case, we are given the initial concentration (C1) as \(6.42 \times 10^{-2} \mathrm{m} M\) and initial volume (V1) as 5.00 mL. We need to find the final concentration (C2) after dilution to a final volume (V2) of 25.0 mL. Again, we can use the dilution formula for this: \(C1 \times V1 = C2 \times V2\) \((6.42 \times 10^{-2} \mathrm{m} M) \times (5.00 \mathrm{mL}) = C2 \times (25.0 \mathrm{mL})\) Now, we can solve for C2: \(C2 = \dfrac{(6.42 \times 10^{-2} \mathrm{m} M) \times (5.00 \mathrm{mL})}{(25.0 \mathrm{mL})}\) \(C2 = 1.284 \times 10^{-2} \mathrm{m} M\) Thus, the final concentration of Zn2+ ions after dilution is \(1.284 \times 10^{-2} \mathrm{m} M\).

Key Concepts

Concentration Calculations

Concentration calculations are a foundation in chemistry, used to determine how much of a solute is present within a given volume of a solution. These calculations help us understand the strength or potency of the solution by knowing how concentrated the solute particles are. Imagine you're adding sugar to your coffee; the more sugar you add, the more concentrated your drink becomes. In chemistry, concentration is typically measured in molarity (M), which refers to the number of moles of solute per liter of solution. This helps us gauge how much reactants we might need in a chemical reaction, or what effects a solution might have in practical applications.

To work out concentration, you can use the simple formula: \[ \text{Concentration (M)} = \frac{\text{Amount of Solute (moles)}}{\text{Volume of Solution (liters)}} \] In practice, you will often need to convert volumes and amounts into moles and liters. This steady approach gives insights into how strong or weak a solution is, preparing you to observe and predict chemical behaviors accurately.

Molarity and Solution Preparation

Molarity is an integral component of solution preparation in chemistry. It's defined as the moles of solute divided by the liters of solution, and it helps chemists to accurately quantify the components of a solution. Preparing a solution with a specific molarity is essential when performing experiments because it ensures consistency and repeatability.

Here's the basic approach to preparing a solution:

• Determine the desired molarity of the solution.
• Calculate the number of moles of solute needed using the formula: \( \text{Moles of solute} = \text{Molarity} \times \text{Volume in liters} \).
• Dissolve the calculated amount of solute in a solvent, usually water, and adjust the total volume to the required amount.

This meticulous method allows for the preparation of solutions with precise concentrations, ensuring that each experimental setup remains consistent.

Dilution Formula

The dilution formula is a handy tool when you need to make a less concentrated solution from a more concentrated one. This is particularly useful when you have limited resources or need specific concentrations for various experiments.

The formula used for dilution is: \[ C_1 \times V_1 = C_2 \times V_2 \] Where:

• \( C_1 \) is the initial concentration of the solution.
• \( V_1 \) is the initial volume of the solution.
• \( C_2 \) is the final concentration after dilution.
• \( V_2 \) is the final total volume after dilution.

To use this formula, balance the product of the initial concentration and volume with the new, diluted concentration and final volume. This equation assumes a conservative system where the quantity of solute remains constant, but the solution volume changes. It's a straightforward approach that allows chemists to determine the new concentration quickly.