Lesson 11.3 Factors Affecting Solubility
Suggested Reading
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Zumdahl Chapter 11 Section11.3
Essential Questions
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What factors affect solubility?
Learning Objectives
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Identify and explain factors affecting solubility.
Predict the relative solubilities of simple molecules based on structure.
Solve problems involving Henry's law.
Introduction
There are three main factors that control solubility of a solute.
1. Temperature
2. Nature of solute or solvent
3. Pressure
In this lesson we will take a look at each.
Temperature
Generally, solubility increases with the rise in temperature and decreases
with the fall of temperature, although as you have seen this is not always
the case. The following rules will help you determine the effects of
temperature.
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In an endothermic process, solubility increases with a rise in
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temperature and vice versa.
In an exothermic process, solubility decreases with a rise in
temperature.
Gases are more soluble in cold solvent than in hot solvent.
Nature of Solute and Solvent
The solubility of a solute in a solvent purely depends on the nature of both
solute and solvent.
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A polar solute dissolves in a polar solvent.
A polar solute has a low solubility in a non-polar solvent.
A non-polar solute dissolves in a non-polar solvent.
A non-polar solute has a low solubility in a polar solvent.
Pressure
In general, pressure change has little effect on the solubility of a liquid or
solid in water, but the solubility of a gas is very much affected by pressure.
There is only one rule to remember.
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An increase in pressure increases the solubility of a gas in a liquid.
For example, carbon dioxide is dissolved in cold cans under pressure.
Henry's Law
The effect of pressure on the solubility of a gas in a liquid can be predicted
quantitatively. According to Henry's law, the solubility of a gas is directly
proportional to the partial pressure of the gas above the solution. Expressed
mathematically, the law is
S = kHP
where S is the solubility of the gas (C is used to denote S in your
textbook)(expressed as mass of solute per unit volume of solvent), kH is
Henry's law constant for the as for a particular liquid at a given temperature,
and P is the partial pressure of the gas. Henry's law can be used to compare
the solubility of a solute under different sets of conditions. For these types of
problems we use the following relation
Lets look at an example.
Example: Applying Henry's Law
27 g of acetylene, C2H2, dissolves in 1 L of acetone at 1.0 atm. In the partial
pressure of acetylene is increased to 12 atm, what is the solubility of acetone.
Solution
At 1.0 atm the partial pressure of acetylene (P1), the solubility (S1) is 27 g of
C2H2 per liter of acetone. For partial pressure of 12 atm (P2), S2 is
Solving for S2 gives: