(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?

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Describe the preparation of cach solution, starting with the anhydrous solute and warer and using the corrcsponding volumetric flask: (a) \(25.0 \mathrm{~mL}\) of \(6.0 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}) ;\) (b) \(1.0 \mathrm{~L}\), of \(0.10 \mathrm{M} \mathrm{BaCl}_{2}\) (aq); (c) \(500 \mathrm{~mL}\) of \(0.0010 \mathrm{M} \mathrm{AgNO}_{3}(\mathrm{aq})\).

What mass (in grams) of anhydrous sohute is needed to prepare each of the following solutions? (a) \(1.0 \mathrm{~L}\) of \(0.10 \mathrm{M} \mathrm{NaCl}(\mathrm{aq}) ;\) (b) \(250 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) \(\mathrm{CaCl}_{2}(\mathrm{aq}) ;\) (c) \(500 \mathrm{~mL}\) of \(0.63 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})\).

In a laboratory exercise, a student mixes \(2.5 .0 \mathrm{~g}\) of cthanol, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\), with \(150 \mathrm{~g}\) of water. (a) What is the mole fraction of ethanol in the solution? (b) What is the molality of cthanol in the solution?

(a) Determine the mass of anhydrous copper(II) sulfate that must be used to prepare \(250 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}(\mathrm{aq})\) solution. (b) Determine the mass of \(\mathrm{CuSO}_{4}-5 \mathrm{H}_{2} \mathrm{O}\) that must be used to prepare \(250 \mathrm{ml}\). of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}(\mathrm{aq})\) solution.

Identify the following as homogencous or hcterogeneous mixtures and suggest a technique for scparating their components: (a) gasolinc and motor oil; (b) carbonated water; (c) charcoal and sugar.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free