Problem 10 - Chemical Principles: The Quest for Insight

Question

(a) A 12.56-mL sample of \(1.345 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is diluted to \(250.0 \mathrm{~mL}\). What is the molar concentration of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the diluted solution? (b) A \(25.00-\mathrm{mL}\) sample of \(0.366 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) is drawn from a reagent bottle with a pipet. The sample is transferred to a \(125.00-\mathrm{mL}\) volumetric flask and diluted to the mark with water. What is the molar concentration of the dilute hydrochloric acid solution?

Step-by-Step Solution

Answer

The final molar concentration of diluted \( K_2SO_4 \) is 0.0676 M and the final molar concentration of diluted \( HCl \) is 0.0732 M.

Step 1: Understand the Dilution Concept

The dilution of a solution involves adding more solvent to decrease the concentration of the solute. The number of moles of solute remains the same before and after dilution. This can be represented by the formula: \( C_1V_1 = C_2V_2 \), where \(C_1\) and \(V_1\) are the initial concentration and volume, and \(C_2\) and \(V_2\) are the final concentration and volume.

Step 2: Calculate the Final Concentration for \( K_2SO_4 \)

Use the dilution formula to find the final concentration of \( K_2SO_4 \). Plug in the initial concentration \(C_1 = 1.345 \, M\), the initial volume \(V_1 = 12.56 \, mL\) and the final volume \(V_2 = 250.0 \, mL\), to find \(C_2\). \[ C_2 = \frac{C_1 \times V_1}{V_2} \] \[ C_2 = \frac{1.345 \, M \times 12.56 \, mL}{250.0 \, mL} \] \[ C_2 = \frac{16.9032 \, mM}{250.0 \, mL} \] \[ C_2 = 0.0676 \, M \] The final concentration of \( K_2SO_4 \) is 0.0676 M.

Step 3: Calculate the Final Concentration for \( HCl \)

Use the dilution formula for the \( HCl \) solution, with \(C_1 = 0.366 \, M\), \(V_1 = 25.00 \, mL\), and \(V_2 = 125.00 \, mL\). \[ C_2 = \frac{C_1 \times V_1}{V_2} \] \[ C_2 = \frac{0.366 \, M \times 25.00 \, mL}{125.00 \, mL} \] \[ C_2 = \frac{9.15 \, mM}{125.00 \, mL} \] \[ C_2 = 0.0732 \, M \] The final concentration of \( HCl \) after dilution is 0.0732 M.

Key Concepts

Molar Concentration

Molar concentration, often represented as molarity and denoted as M, is a measure of the concentration of a solute in a solution in terms of amount of substance per unit volume. To put it plainly, it tells us how many moles of solute are present in one liter of solution. Understanding molar concentration is fundamental in chemistry as it allows you to predict the outcome of reactions and prepare solutions with precise chemical concentrations.

For instance, a solution with a molarity of 1.345 M contains 1.345 moles of solute in each liter. This concept simplifies dealing with chemical reactions since quantities can directly relate to the volume of solutions used.

Dilution Formula

The dilution formula is a straightforward equation used to calculate the resulting molar concentration of a solution after it has been diluted with additional solvent. It is given as: \( C_1V_1 = C_2V_2 \), where \(C_1\) is the initial concentration, \(V_1\) is the initial volume, \(C_2\) is the final concentration, and \(V_2\) is the final volume after dilution.

The power of the dilution formula lies in its reflection of the fact that the number of moles of solute remains constant during the dilution process. As the volume increases, the same amount of substance spreads out into a larger space, consequently reducing its concentration. Applying this formula not only confirms our understanding of conservation of mass but also ensures precision in creating solutions of desired molarity.

Moles of Solute

Moles of solute is a key term in chemistry that symbolizes the actual amount of a substance present in a solution. One mole is defined as exactly 6.02214076×10²³ particles (atoms, molecules, ions, or electrons), known as Avogadro's number. This concept allows chemists to count particles in a given mass of a substance using its molar mass.

The number of moles of solute is crucial when it comes to preparing solutions and conducting chemical reactions. In the case of dilution, although the concentration changes, the moles of the solute in the solution remain the same since no additional solute is added or removed -- the volume simply changes. The calculation of moles is fundamental to applying the dilution formula correctly and integral to understanding the changes occurring during the dilution process.

Chemistry Education

Chemistry education provides the knowledge and skills necessary to understand, interpret, and manipulate chemical information. A solid foundation in chemistry can enable students to solve practical problems and make informed decisions in everyday life. Key to effective chemistry education is breaking down complex topics, like molar concentration and dilution, into more digestible concepts for students. By using real-life examples, interactive experiments, and clear explanations, educators can stimulate curiosity and enhance the learning experience.

In our textbook exercise, providing a step-by-step solution allows students to follow the logic and methods used to solve chemistry problems. Educational strategies such as these are significant because they not only help students with their homework but also foster a deeper understanding of chemical principles.