Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

How do the lime-soda and ion-exchange processes differ in treating hard water?

Short Answer

Expert verified
The lime-soda process precipitates hard ions, while ion-exchange replaces them with sodium ions, suited for different scales and outputs.

Step by step solution

01

Understanding Lime-Soda Process

The lime-soda process involves adding lime (Ca(OH)₂) and soda ash (Na₂CO₃) to hard water. The lime is used to remove calcium hardness by precipitating calcium carbonate (CaCO₃), while soda ash is used to remove magnesium hardness by precipitating magnesium hydroxide (Mg(OH)₂). The precipitates are then removed via sedimentation and filtration, leaving softened water.
02

Understanding Ion-Exchange Process

The ion-exchange process uses resin beads charged with sodium ions (Na⁺) to replace calcium and magnesium ions present in hard water. As hard water passes through the resin, calcium (Ca²⁺) and magnesium (Mg²⁺) ions are exchanged with sodium ions, softening the water. The resin is periodically regenerated using a concentrated sodium chloride solution (brine).
03

Comparing Chemical Reactions

In the lime-soda process, chemical reactions involve precipitation, which physically removes hardness causing minerals. In contrast, the ion-exchange process relies on a chemical exchange between ions without precipitating them.
04

Operational Differences

The lime-soda process is more suited for large-scale water treatment plants and requires handling of chemicals and resultant sludge. The ion-exchange process is typically used in household water softeners and involves regenerating the resin with brine, making it less suited for large-scale operations.
05

Efficiency and Output

The lime-soda process can effectively remove both calcium and magnesium ions, but requires large apparatus for sedimentation and filtration. The ion-exchange process efficiently softens water but leaves sodium ions, which may not be desirable in all applications.

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.

Lime-Soda Process
The lime-soda process is a traditional method for treating hard water. This technique adds lime (Ca(OH)₂) and soda ash (Na₂CO₃) to water. Each chemical targets specific minerals responsible for water hardness.

In this process:
  • Lime (Ca(OH)₂): It helps to remove calcium hardness by forming calcium carbonate (CaCO₃) precipitates.
  • Soda Ash (Na₂CO₃): It targets magnesium hardness, creating magnesium hydroxide (Mg(OH)₂) precipitates.
These precipitates are then separated from the water through sedimentation and filtration.

Advantages:
  • Effectively removes both calcium and magnesium, reducing the total hardness of water.
  • Suitable for large-scale operations, such as municipal water treatment plants.
Drawbacks:
  • Produces sludge that must be managed and disposed of properly.
  • Requires significant equipment and space for sedimentation and filtration.
Ion-Exchange Process
The ion-exchange process is a modern and efficient method for water softening, often used in residential setups.

This method involves resin beads charged with sodium (Na⁺) ions, which are crucial for the exchange of calcium (Ca²⁺) and magnesium (Mg²⁺) ions found in hard water.
When hard water flows over these resin beads, an exchange reaction occurs:
  • Calcium and magnesium ions clamp onto the resin and displace the sodium ions.
  • As a result, the softened water carries the lesser reactive sodium ions.
Key Features:
  • Efficient in replacing hard minerals and providing soft water instantly.
  • Resin beads can be regenerated using a saline solution, restoring their functionality.

Considerations:
  • May increase sodium content in the treated water, affecting its taste or suitability for low-sodium diets.
  • Regular maintenance required to regenerate the resin beads.
Calcium and Magnesium Removal
Calcium and magnesium ions are the main contributors to water hardness. Their removal is crucial for water softening.

Calcium Removal:
  • In the lime-soda process, lime helps in converting calcium ions into insoluble calcium carbonate precipitates.
  • During ion-exchange, calcium ions are exchanged with sodium ions on the resin beads.

Magnesium Removal:
  • In the lime-soda process, soda ash helps in transforming magnesium ions into magnesium hydroxide, which is removed as a solid.
  • Similarly, in ion-exchange, magnesium ions swap places with sodium ions on the resin.
These methods ensure the reduction of hard water minerals, making water softer and better suited for everyday use.
Water Softening Methods
Water softening is essential to combat the effects of hard water, such as scale buildup and reduced soap effectiveness.

The two main water softening methods discussed, the lime-soda and the ion-exchange processes, offer distinct advantages depending on the scale and usage context.
Lime-Soda Process:
  • Best suited for larger installations where large quantities of water need treatment.
  • Effective but requires significant infrastructure and handling of chemical byproducts.

Ion-Exchange Process:
  • Common in residential water treatment systems due to its compactness and ease of use.
  • Requires regular maintenance to keep the resin active and efficient.
In choosing a method, one must consider effectiveness, maintenance needs, and the intended use of the softened water. Applying the correct strategy ensures better water quality while remaining efficient and practical for the intended application.

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

Consider a \(13.0 \%\) solution of sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), whose density is \(1.090 \mathrm{~g} / \mathrm{mL}\). (a) Calculate the molarity of this solution. (b) To what volume should \(100 . \mathrm{mL}\) of this solution be diluted to prepare a 1.10 -M solution?

Which of these general types of substances would you expect to dissolve readily in water? Explain why. (a) Alcohols (b) Hydrocarbons (c) Metals (d) Nonpolar molecules (e) Polar molecules (f) Salts

What is the difference between solubility and miscibility?

The partial pressure of \(\mathrm{O}_{2}\) in your lungs is about \(100 \mathrm{mmHg} .\) Calculate the concentration of \(\mathrm{O}_{2}\) (in grams per liter) that can dissolve in water at \(37^{\circ} \mathrm{C}\) when the \(\mathrm{O}_{2}\) partial pressure is \(100 . \mathrm{mmHg} .\) The Henry's law constant for \(\mathrm{O}_{2}\) at \(37{ }^{\circ} \mathrm{C}\) is \(1.5 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{mmHg}^{-1}\).

A \(0.63 \%\) by weight aqueous tin(II) fluoride, \(\mathrm{SnF}_{2},\) solution is used as an oral rinse in dentistry to decrease tooth decay. (a) Calculate the \(\mathrm{SnF}_{2}\) concentration in ppm and ppb. (b) Calculate the molarity of \(\mathrm{SnF}_{2}\) in the solution. (c) Tin(II) fluoride is produced commercially by a series of steps that begin with reduction of cassiterite, \(\mathrm{SnO}_{2}\), the principal tin ore, with carbon.$$\mathrm{SnO}_{2}(\mathrm{~s})+2 \mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{Sn}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g})$$ Once purified, the tin is reacted with hydrogen fluoride vapor to produce \(\mathrm{SnF}_{2}\). Consider that one metric ton of cassiterite ore is reduced with sufficient carbon to tin metal in \(80 \%\) yield. The tin is purified and reacted with sufficient hydrogen fluoride to produce \(\mathrm{SnF}_{2}\) in \(94 \%\) yield. Calculate how many \(250 .-\mathrm{mL}\) bottles of \(0.63 \%\) \(\mathrm{SnF}_{2}\) solution could be prepared from the one metric ton of cassiterite by these steps.

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