1. Christopher G. Hamaker, Illinois State University, Normal IL
Introductory Chemistry: Concepts & Connections 4th Edition by Charles H. Corwin
Chapter 14
Solutions
Christopher G. Hamaker, Illinois State University, Normal IL
© 2005, Prentice Hall

2. Solutions A solution is a homogeneous mixture.
A solution is composed of a solute dissolved in a solvent.
Solutions exist in all three physical states:
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3. Gases in Solution Temperature effects the solubility of gases.
The higher the temperature, the lower the solubility of a gas in solution.
An example is carbon dioxide in soda:
Less CO2 escapes when you open a cold soda than when you open the soda warm
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4. Pressure & Gas Solubility
Pressure also influences the solubility of gases.
According to Henry’s Law, the solubility of a gas is directly proportional to the partial pressure of the gas above the liquid.
If we double the partial pressure of a gas, we double the solubility.
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5. Henry’s Law
We can calculate the solubility of a gas at an new pressure using Henry’s Law.
= new solubility
solubility ×
new pressure
old pressure
What is the solubility of oxygen gas at 25C and a partial pressure of 1150 torr if the solubility of oxygen is g/100 mL at 25C and 760 torr?
1150 torr
760 torr
= g/100 mL
g/100 mL ×
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6. Polar Molecules
When two liquids make a solution, the solute is the lesser quantity, and the solvent is the greater quantity.
Recall, that a net dipole is present in a polar molecule.
Water is a polar molecule.
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7. Polar & Nonpolar Solvents
A liquid composed of polar molecules is a polar solvent. Water and ethanol are polar solvents.
A liquid composed of nonpolar molecules is a nonpolar solvent. Hexane is a nonpolar solvent.
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8. Like Dissolves Like Polar solvents dissolve in one another.
Nonpolar solvents dissolve in one another.
This it the like dissolves like rule.
Methanol dissolves in water but hexane does not dissolve in water.
Hexane dissolves in toluene, but water does not dissolve in toluene.
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9. Miscible & Immiscible
Two liquids that completely dissolve in each other are miscible liquids.
Two liquids that are not miscible in each other are immiscible liquids.
Polar water and nonpolar oil are immiscible liquids and do not mix to form a solution.
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10. Solids in Solution
When a solid substance dissolves in a liquid, the solute particles are attracted to the solvent particles.
When a solution forms, the solute particles are more strongly attracted to the solvent particles than other solute particles.
We can also predict whether a solid will dissolve in a liquid by applying the like dissolves like rule.
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11. Like Dissolves Like for Solids
Ionic compounds, like sodium chloride, are soluble in polar solvents and insoluble in nonpolar solvents.
Polar compounds, like table sugar (C12H22O11), are soluble in polar solvents and insoluble in nonpolar solvents.
Nonpolar compounds, like naphthalene (C10H8), are soluble in nonpolar solvents and insoluble in polar solvents.
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12. The Dissolving Process
When a soluble crystal is placed into a solvent, it begins to dissolve.
When a sugar crystal is placed in water, the water molecules attack the crystal and begin pulling part of it away and into solution.
The sugar molecules are held within a cluster of water molecules called a solvent cage.
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13. Dissolving of Ionic Compounds
When a sodium chloride crystal is place in water, the water molecules attack the edge of the crystal.
In an ionic compound, the water molecules pull individual ions off of the crystal.
The anions are surrounded by the positively charged hydrogens on water.
The cations are surrounded by the negatively charged oxygen on water.
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14. Rate of Dissolving
There are three ways we can speed up the rate of dissolving for a solid compound.
Heating the solution:
This increases the kinetic energy of the solvent and the solute is attacked faster by the solvent molecules.
Stirring the solution:
This increases the interaction between solvent and solute molecules.
Grinding the solid solute:
There is more surface area for the solvent to attack.
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15. Solubility and Temperature
The solubility of a compound is the maximum amount of solute that can dissolve in 100 g of water at a given temperature.
In general, a compound becomes more soluble as the temperature increases.
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16. Saturated Solutions
A solution containing exactly the maximum amount of solute at a given temperature is a saturated solution.
A solution that contains less than the maximum amount of solute is an unsaturated solution.
Under certain conditions, it is possible to exceed the maximum solubility of a compound. A solution with greater than the maximum amount of solute is a supersaturated solution.
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17. Supersaturated Solutions
At 55C, the solubility of NaC2H3O2 is 100 g per 100 g water.
If a saturated solution at 55C is cooled to 20C, the solution is supersaturated.
Supersaturated solutions are unstable. The excess solute can readily be precipitated.
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18. Supersaturation
A single crystal of sodium acetate added to a supersaturated solution of sodium acetate in water causes the excess solute to rapidly crystallize from the solution.
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19. Concentration of Solutions
The concentration of a solution tells us how much solute is dissolved in a given quantity of solution.
We often hear imprecise terms such as a “dilute solution” or a “concentrated solution”.
There are two precise ways to express the concentration of a solution:
mass/mass percent
molarity
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20. Mass Percent Concentration
Mass percent concentration compares the mass of solute to the mass of solvent.
The mass/mass percent (m/m %) concentration is the mass of solute dissolved in 100 g of solution.
mass of solute
mass of solution
× 100% = m/m %
g solute
g solute + g solvent
× 100% = m/m %
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21. Calculating Mass/Mass Percent
A student prepares a solution from 5.00 g NaCl dissolved in 97.0 g of water. What is the concentration in m/m %?
5.50 g NaCl
5.00 g NaCl g H2O
× 100% = m/m %
5.00 g NaCl
102 g solution
× 100% = %
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22. Mass Percent Unit Factors
We can write several unit factors based on the concentration 4.90 m/m% NaCl:
4.90 g NaCl
100 g solution
4.90 g NaCl
95.1 g water
95.1 g water
100 g solution
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23. Mass Percent Calculation
What mass of a 5.00 m/m% solution of sucrose contains 25.0 grams of sucrose?
We want grams solution, we have grams sucrose.
100 g solution
5.00 g sucrose
= 500 g solution
25.0 g sucrose ×
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24. Molar Concentration
The molar concentration, or molarity (M), is the number of moles of solute per liter of solution, is expressed as moles/liter.
Molarity is the most commonly used unit of concentration.
moles of solute
liters of solution
= M
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25. Calculating Molarity
What is the molarity of a solution containing g of NaOH in L of solution?
We also need to convert grams NaOH to moles NaOH (MM = g/mol).
= M NaOH ×
18.0 g NaOH
0.100 L solution
1 mol NaOH
40.00 g NaOH
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26. Molarity Unit Factors
We can write several unit factors based on the concentration 4.50 M NaOH:
4.50 mol NaOH
1 L solution
4.50 mol NaOH
1000 mL solution
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27. Molar Concentration Problem
How many grams of K2Cr2O7 are in mL of M K2Cr2O7?
We want mass K2Cr2O7, we have mL solution.
0.100 mol K2Cr2O7
1000 mL solution
250.0 mL solution × ×
294.2 g K2Cr2O7
1 mol K2Cr2O7
= g K2Cr2O7
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28. Molar Concentration Problem
What volume of 12.0 M HCl contains 7.30 g of HCl solute (MM = g/mol)?
We want volume, we have grams HCl.
1 mol HCl
36.46 g HCl
7.30 g HCl × ×
1000 mL solution
12.0 mol HCl
= mL solution
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29. Dilution of a Solution
Rather than prepare a solution by dissolving a solid in water, we can prepare a solution by diluting a more concentrated solution.
When performing a dilution, the amount of solute does not change, only the amount of solvent.
The equation we use is: M1 × V1 = M2 × V2
M1 and V1 are the initial molarity and volume and M2 and V2 are the new molarity and volume
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30. Dilution Problem
What volume of 6.0 M NaOH needs to be diluted to prepare 5.00 L if 0.10 M NaOH?
We want final volume and we have our final volume and concentration.
M1 × V1 = M2 × V2
(6.0 M) × V1 = (0.10 M) × (5.00 L)
V1 =
= L
(0.10 M) × (5.00 L)
6.0 M
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31. Solution Stoichiometry
In Chapter 10, we performed mole calculations involving chemical equations, stoichiometry problems.
We can also apply stoichiometry calculations to solutions.
molarity known  moles known 
moles unknown  mass unknown
solution
concentration
balanced
equation
molar mass
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32. Solution Stoichiometry Problem
What mass of silver bromide is produced from the reaction of 37.5 mL of M aluminum bromide with excess silver nitrate solution?
AlBr3(aq) + 3 AgNO3(aq) → 3 AgBr(s) + Al(NO3)3(aq)
We want g AgBr, we have volume of AlBr3
37.5 mL soln ×
3 mol AgBr
1 mol AlBr3
0.100 mol AlBr3
1000 mL soln ×
1 mol AgBr
g AgBr
= 2.11 g AgBr
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33. Conclusions Gas solubility decreases as the temperature increases.
Gas solubility increases as the pressure increases.
When determining whether a substance will be soluble in a given solvent, apply the like dissolves like rule.
Polar molecules dissolve in polar solvents.
Nonpolar molecules dissolved in nonpolar solvents.
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34. Conclusions Continued
Three factors can increase the rate of dissolving for a solute:
Heating the solution
Stirring the solution
Grinding the solid solute
In general, the solubility of a solid solute increases as the temperature increases.
A saturated solution contains the maximum amount of solute at a given temperature.
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35. Conclusions Continued
The mass/mass percent concentration is the mass of solute per 100 grams of solution:
The molarity of a solution is the moles of solute per liter of solution.
mass of solute
mass of solution
× 100% = m/m %
moles of solute
liters of solution
= M
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36. Conclusions Continued
You can make a solution by diluting a more concentrated solution:
M1 × V1 = M2 × V2
We can apply stoichiometry to reactions involving solutions using the molarity as a unit factor to convert between moles and volume.
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