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By the end of this module, you will be able to:
  • Describe the effects of temperature and pressure on solubility
  • State Henry’s law and use it in calculations involving the solubility of a gas in a liquid
  • Explain the degrees of solubility possible for liquid-liquid solutions

Imagine adding a small amount of salt to a glass of water, stirring until all the salt has dissolved, and then adding a bit more. You can repeat this process until the salt concentration of the solution reaches its natural limit, a limit determined primarily by the relative strengths of the solute-solute, solute-solvent, and solvent-solvent attractive forces discussed in the previous two modules of this chapter. You can be certain that you have reached this limit because, no matter how long you stir the solution, undissolved salt remains. The concentration of salt in the solution at this point is known as its solubility.

The solubility    of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium . Referring to the example of salt in water:

NaCl ( s ) Na + ( a q ) + Cl ( a q )

When a solute’s concentration is equal to its solubility, the solution is said to be saturated    with that solute. If the solute’s concentration is less than its solubility, the solution is said to be unsaturated    . A solution that contains a relatively low concentration of solute is called dilute, and one with a relatively high concentration is called concentrated.

If we add more salt to a saturated solution of salt, we see it fall to the bottom and no more seems to dissolve. In fact, the added salt does dissolve, as represented by the forward direction of the dissolution equation. Accompanying this process, dissolved salt will precipitate, as depicted by the reverse direction of the equation. The system is said to be at equilibrium when these two reciprocal processes are occurring at equal rates, and so the amount of undissolved and dissolved salt remains constant. Support for the simultaneous occurrence of the dissolution and precipitation processes is provided by noting that the number and sizes of the undissolved salt crystals will change over time, though their combined mass will remain the same.

Solutions may be prepared in which a solute concentration exceeds its solubility. Such solutions are said to be supersaturated    , and they are interesting examples of nonequilibrium states. For example, the carbonated beverage in an open container that has not yet “gone flat” is supersaturated with carbon dioxide gas; given time, the CO 2 concentration will decrease until it reaches its equilibrium value.

Solutions of gases in liquids

In an earlier module of this chapter, the effect of intermolecular attractive forces on solution formation was discussed. The chemical structures of the solute and solvent dictate the types of forces possible and, consequently, are important factors in determining solubility. For example, under similar conditions, the water solubility of oxygen is approximately three times greater than that of helium, but 100 times less than the solubility of chloromethane, CHCl 3 . Considering the role of the solvent’s chemical structure, note that the solubility of oxygen in the liquid hydrocarbon hexane, C 6 H 14 , is approximately 20 times greater than it is in water.

Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
Syamthanda Reply
hey , can you please explain oxidation reaction & redox ?
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hey , can you please explain oxidation reaction and redox ?
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for grade 12 or grade 11?
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the value of V1 and V2
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advantages of electrons in a circuit
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it is the force or component of the force that the surface exert on an object incontact with it and which acts perpendicular to the surface
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what is the half reaction of Potassium and chlorine
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a structure of a thermocouple used to measure inner temperature
Anna Reply
a fixed gas of a mass is held at standard pressure temperature of 15 degrees Celsius .Calculate the temperature of the gas in Celsius if the pressure is changed to 2×10 to the power 4
Amahle Reply
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a rate of change in velocity of an object whith respect to time
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Acceleration is a rate of change in velocity.
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t =r×f
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hi
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Leago
use fnet method. how many obects are being calculated ?
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Lungile Reply
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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