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Chapter 11: Problem 99

What stabilizes a colloidal suspension? Explain why adding heat or adding an electrolyte can cause the suspended particles to settle out

Short Answer

Expert verified
A colloidal suspension is stabilized by the balance of repulsive and attractive forces between the dispersed particles. Repulsive forces, mainly due to the electrical double layer and steric hindrance, prevent aggregation, while attractive forces, such as van der Waals forces, can cause aggregates or precipitates if they dominate. Adding heat increases the kinetic energy and Brownian motion of particles, making them more likely to collide and aggregate. Adding an electrolyte reduces the electrical double layer, decreasing repulsive forces and allowing attractive forces to dominate, leading to aggregation and sedimentation.

Step by step solution

01

Understand colloidal suspensions

A colloidal suspension is a mixture of insoluble particles (solid, liquid, or gas) dispersed in a continuous medium (such as a liquid). The particles have dimensions in the range of 1 to 1000 nanometers. The stability of a colloidal suspension depends on the balance of repulsive and attractive forces between the dispersed particles. The repulsive forces prevent the particles from approaching each other and aggregating, while the attractive forces can cause aggregates or precipitates if they dominate.
02

Examine repulsive forces in colloidal suspension

In many colloids, the repulsive forces are mainly due to the electrical double layer around the particles. This double layer consists of a charged surface (either positive or negative) and a region of counter ions surrounding the surface. The net effect of this double layer is that two neighboring particles will experience repulsive electrostatic forces when they approach each other. The intervening liquid also contributes to the repulsive forces through steric hindrance (i.e., the entropic penalty associated with forcing solvent molecules out of the space between particles).
03

Examine attractive forces in colloidal suspension

The primary attractive forces in colloidal suspensions are van der Waals forces, which arise due to attractive interactions between electron clouds of particles. These forces can cause aggregation or precipitation if they are not counterbalanced by repulsive forces.
04

Effect of heat on colloidal suspension

Adding heat to a colloidal suspension can cause the suspended particles to settle out due to an increase in Brownian motion. When the temperature of the suspension increases, the particles gain kinetic energy and move more rapidly. The enhanced particle motion means that the particles are more likely to collide with each other and overcome the repulsive forces, leading to aggregation or sedimentation of the particles.
05

Effect of electrolyte on colloidal suspension

Adding an electrolyte to a colloidal suspension can also cause the suspended particles to settle out due to a reduction in the electrical double layer and the consequent decrease in repulsive electrostatic forces. The added electrolyte ions will interact with the charged surfaces of the particles and compress the electrical double layer, effectively screening the surface charge. This reduces the repulsive forces between particles and allows the attractive van der Waals forces to dominate, leading to particle aggregation and sedimentation.

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Most popular questions from this chapter

In a coffee-cup calorimeter, 1.60 \(\mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) was mixed with 75.0 \(\mathrm{g}\) water at an initial temperature \(25.00^{\circ} \mathrm{C}\) . After dissolution of the salt, the final temperature of the calorimeter contents was \(23.34^{\circ} \mathrm{C}\) . a. Assuming the solution has a heat capacity of 4.18 \(\mathrm{J} / \mathrm{g}\) \(^{\circ} \mathrm{C},\) and assuming no heat loss to the calorimeter, calculate the enthalpy of solution \(\left(\Delta H_{\mathrm{soln}}\right)\) for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of $\mathrm{kJ} / \mathrm{mol} .$ b. If the enthalpy of hydration for \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is $-630 . \mathrm{kJ} / \mathrm{mol}\( calculate the lattice energy of \)\mathrm{NH}_{4} \mathrm{NO}_{3} .$

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