Problem 158
Question
Which of the following statements are correct? (a) Presence of equal and similar charges on colloidal particles provides stability to the colloids (b) Mixing two oppositely chartged sols neutralises their charges and stabilises the colloid. (c) Brownian movement stabilises sols (d) Any amount of dispersed liquid can be added to emulsion without destabilising it.
Step-by-Step Solution
Verified Answer
Statements (a) and (c) are correct.
1Step 1: Analyzing Statement (a)
Statement (a) claims that the presence of equal and similar charges on colloidal particles provides stability to the colloids. This is true because the like charges on colloidal particles repel each other, preventing them from coming together and aggregating, thus providing stability to the colloid.
2Step 2: Evaluating Statement (b)
Statement (b) suggests that mixing two oppositely charged sols neutralizes their charges and stabilizes the colloid. This is incorrect because neutralizing the charges often leads to coagulation or aggregation of particles, decreasing the stability of the colloid.
3Step 3: Assessing Statement (c)
Statement (c) indicates that Brownian movement stabilizes sols. This is correct as Brownian movement causes constant motion of the particles, preventing them from settling down and hence maintaining dispersion stability.
4Step 4: Reviewing Statement (d)
Statement (d) asserts that any amount of dispersed liquid can be added to an emulsion without destabilizing it. This is false because adding too much dispersed phase can exceed the emulsion's capacity and cause phase separation, leading to destabilization.
Key Concepts
Brownian MovementColloidal ParticlesOppositely Charged Sols
Brownian Movement
Brownian movement is a fascinating phenomenon observed in colloidal systems, named after the botanist Robert Brown who first documented it. This movement refers to the random, zigzagging motion of colloidal particles suspended in a fluid.
But why do these tiny particles dance around so unpredictably?
It's because they are continuously bombarded by the much smaller, faster-moving molecules of the dispersing medium, typically a liquid or gas. This bombardment keeps the colloidal particles in relentless motion.
Brownian movement plays a crucial role in stabilizing colloidal systems. This ceaseless movement prevents particles from settling under the influence of gravity, helping maintain a uniform dispersion of particles throughout the solvent. Here are a few key points about Brownian movement:
But why do these tiny particles dance around so unpredictably?
It's because they are continuously bombarded by the much smaller, faster-moving molecules of the dispersing medium, typically a liquid or gas. This bombardment keeps the colloidal particles in relentless motion.
Brownian movement plays a crucial role in stabilizing colloidal systems. This ceaseless movement prevents particles from settling under the influence of gravity, helping maintain a uniform dispersion of particles throughout the solvent. Here are a few key points about Brownian movement:
- It is more pronounced in smaller particles, as their mass is not enough to counteract the molecular collisions effectively.
- This movement contributes significantly to the stability of sols, as it impedes sedimentation.
- It is a temperature-dependent process; higher temperatures increase molecular motion, thus enhancing Brownian movement.
Colloidal Particles
Colloidal particles are small, but they play a huge role in the stability and functionality of colloids. They are particles that range from 1 to 1000 nanometers in size and can exist in various states such as solid, liquid, or gas, distributed within another medium.
These particles don't settle out of the dispersing medium in a reasonable timeframe due to their small size and light weight. One of the primary reasons colloidal particles remain dispersed is their surface charges. Most colloidal particles carry an electrical charge.
When uniformly charged, these particles experience mutual repulsion, which keeps them from clumping together.
These particles don't settle out of the dispersing medium in a reasonable timeframe due to their small size and light weight. One of the primary reasons colloidal particles remain dispersed is their surface charges. Most colloidal particles carry an electrical charge.
When uniformly charged, these particles experience mutual repulsion, which keeps them from clumping together.
- The nature of colloidal particles, including their charge and size, strongly affects the colloid's properties.
- The electric charge density on the surface of these particles determines their repulsive forces.
- In many colloids, stabilization is achieved by adsorption of ions onto the particle surface, enhancing the system's stability.
Oppositely Charged Sols
Imagine a scenario where two different colloidal sols, each carrying an opposite charge, come into contact. This leads to interesting interactions between them.
When oppositely charged sols mix, the positive and negative charges can neutralize each other, potentially leading to the destabilization of the colloid.
This happens because the reduction in surface charges diminishes the repulsive forces keeping the particles apart, thus allowing them to aggregate or coalesce.
When oppositely charged sols mix, the positive and negative charges can neutralize each other, potentially leading to the destabilization of the colloid.
This happens because the reduction in surface charges diminishes the repulsive forces keeping the particles apart, thus allowing them to aggregate or coalesce.
- If complete neutralization occurs, the colloidal particles may clump together, removing them from the dispersion medium in a process known as coagulation.
- Neutralizing oppositely charged sols often results in precipitation as the now weakly charged or neutral particles attract, stick together, and form larger agglomerates.
- This is a critical point in the preparation and handling of colloids, as maintaining the appropriate charge balance is key to preserving their stability.
Other exercises in this chapter
Problem 156
Freundlich adsorption isotherm is given by expression \(\frac{x}{m}=k p^{\frac{1}{n}}\) which of the following conclusions can be drawing from this expression (
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\(\mathrm{H}_{2}\) gas is adsorbed on activated charcoal to a very little extend in comparison to easily liquefiable gases due to (a) Very strong van der Waal's
View solution Problem 159
An emulsion cannot be broken by and (a) Adding more amount of dispersion medium (b) Heating (c) Adding emulsifying agent (d) Freezing
View solution Problem 160
Which of the following substances will precipitate the negatively changed emulsions? (a) Glucose (b) Urea (c) \(\mathrm{NaCl}\) (d) \(\mathrm{KCl}\)
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