1. 14 Solutions Brass, a solid solution of Zn and Cu, is used to make
musical instruments and many other objects.
Foundations of College Chemistry, 14th Ed.
Morris Hein and Susan Arena
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2. © 2014 John Wiley & Sons, Inc. All rights reserved.
Chapter Outline
14.1 General Properties of Solutions
14.2 Solubility
14.3 Rate of Dissolving Solids
14.4 Concentration of Solutions
14.5 Colligative Properties of Solutions
14.6 Osmosis and Osmotic Pressure
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3. General Properties of Solutions
Solution: a homogeneous mixture of one or more
solutes and a solvent.
Solute: substance being dissolved.
Solvent: dissolving agent that is usually the
most abundant substance in the mixture.
Note: a solution does not always just refer to liquids.
Example: Air is a solution composed of N2, O2, Ar and CO2
N2 is the solvent as it composes 78% of air.
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4. Solutions Practice Soda is a mixture of sugar in water.
Which substance is the solute?
a. sugar
b. water
c. soda
A solution is prepared by adding 25 mL of ethyl alcohol
to 75 mL of water. Which substance is the solvent?
a. ethyl alcohol
b. water
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5. Common Types of Solutions
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6. Properties of a True Solution
A homogeneous mixture of two or more components
whose ratio can be varied.
2. The dissolved solute is molecule or ionic in size
(< 1 nm).
Can be colored or colorless, though solutions are
usually transparent.
4. The solute remains dissolved and does not settle
(precipitate) out of solution over time.
The solute can be separated from solvent by physical
means (usually evaporation).
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7. Solubility Solubility: the amount of a substance that will dissolve
in a specific amount of solvent at a given temperature.
Example g KBr/100g H2O at 23 ºC
Miscible: when two liquids dissolve in each other.
Immiscible: when two liquids do not dissolve one another.
A mixture of oil and water is immiscible.
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8. © 2014 John Wiley & Sons, Inc. All rights reserved.
Solubility Rules
Soluble
Insoluble
Na+, K+, NH4+
Nitrates (NO3-)
Acetates, (C2H3O2-)
Except
Cl-, Br-, I-
Ag+, Hg22+, Pb2+
Sulfates (SO42-),
Ag+, Ca2+ (slightly)
Except
Ba2+, Sr2+, Pb2+
Carbonates (CO32-)
Phosphates (PO43-)
OH-, Sulfides (S2-)
Except
NH4+, Group I
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9. Solutions Practice Predict the solubility of barium sulfate.
a. soluble
b. insoluble
Most sulfates are soluble, except Ba2+.
Predict the solubility of NaCl.
a. soluble
b. insoluble
All Na+ salts are soluble.
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10. Solutions Practice Predict the solubility of silver nitrate.
a. soluble
b. insoluble
All NO3- salts are soluble.
Predict the solubility of silver hydroxide.
a. soluble
b. insoluble
Most hydroxides are insoluble.
Predict the solubility of ammonium carbonate.
a. soluble
b. insoluble
All NH4+ salts are soluble.
© 2014 John Wiley & Sons, Inc. All rights reserved.

11. Factors Affecting Solubility
“Like Dissolves Like”
Polar compounds dissolve in polar solvents.
Ethanol (CH3OH) dissolves in water (HOH).
Nonpolar compounds dissolve in nonpolar solvents.
Carbon tetrachloride (CCl4) dissolves in
hexanes (CH3(CH2)4CH3).
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12. Factors Affecting Solubility
Ionic Compound Solubility in Polar Solvents
Several ionic compounds dissolve in water,
due to strong ion-dipole forces.
The individual cations and anions are surrounded
by H2O molecules (i.e., hydrated).
The cation is attracted to the partially negative O atom.
The anion is attracted to the partially positive H atoms.
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13. Temperature and Solubility
Solubility increases with temperature for most solids (red lines)
Solubility decreases with temperature for all gases (blue lines).
As a gas increases in temperature, the kinetic energy
increases, which means it interacts less with the liquid,
making it less easy to solvate.
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14. Pressure and Solubility
Pressure does not affect solubility of liquids or solids.
Gas solubility in a liquid is proportional to
the gas pressure over the liquid.
: A bottle of root beer is under high pressure.
Example
As the bottle opens, the pressure decreases,
and the bubbles formed indicate gas loss from the liquid.
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15. Saturated and Unsaturated Solutions
There are limits to the solubility of a compound
at a given temperature.
Saturated solutions: contain the maximum amount
of dissolved solute in a solvent.
Saturated solutions are still dynamic; dissolved solute
is in equilibrium with undissolved solute.
undissolved solute dissolved solute
Unsaturated solutions: contain less than the maximum
amount of possible dissolved solute in a solvent.
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16. Supersaturated Solutions
Supersaturated solutions: contain more solute than
needed to saturate a solution at a given temperature.
How is this possible?
Heating a solution can allow more to dissolve.
Upon cooling to ambient temperature,
the solution is supersaturated.
These solutions are unstable -- disturbing the solutions
can cause precipitation of solute.
Some hotpacks release heat by crystallization
of a supersaturated solution of sodium acetate.
© 2014 John Wiley & Sons, Inc. All rights reserved.

17. Solubility Practice Will a solution prepared by adding 9.0 g of KCl to
20.0 g of H2O be saturated or unsaturated at 20 ºC?
Using Table 14.3, 34.0 g of KCl will dissolve
in g of H2O at 20 ºC.
6.8 g of KCl will then dissolve in
20.0 g of water at that temperature.
The KCl solution should be saturated.
© 2014 John Wiley & Sons, Inc. All rights reserved.

18. Rate of Dissolving Solids
Effect of Particle Size
A solid can only dissolve at a surface that
is in contact with the solvent.
Since smaller crystals have a higher surface to volume
area, smaller crystals dissolve faster than larger ones.
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19. Rate of Dissolving Solids
Effect of Temperature
Increasing the temperature normally increases the rate
of dissolution of most compounds.
Solvent molecules strike the solid surface more often,
causing the solid to dissolve more rapidly.
The solute molecules are more easily separate from the solid due to a higher kinetic energy.
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20. Rate of Dissolving Solids
Effect of Solute Concentration
Rate is highest at higher concentration and
decreases at lower concentration.
As the solution approaches the saturation point,
the rate of solute dissolving decreases.
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21. Rate of Dissolving Solids
Effect of Agitation/Stirring
Stirring a solution briskly breaks up a solid
into smaller pieces, increasing surface area,
thereby increasing the rate of dissolution.
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22. Surface Area Surface Area of Two Crystals
Surface area = 6(side)2 = 6(0.1)2 = 0.06 cm2
1000 cubes have a total surface area of
1000 x 0.06 cm2 = 60 cm2
What is the surface area of a 1 cm square crystal?
Surface area = 6(side)2 = 6(1)2 = 6 cm2
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23. Solutions: A Reaction Medium
The purpose of dissolving reactants in a solution is
often to allow them to come in close contact to react.
Example:
Solid-solid reactions are generally very slow
at ambient temperature
KCl (s) + AgNO3 (s) No Reaction
By dissolving both compounds in water,
the ions can collide with one another and react
to form an insoluble compound.
KCl (aq) + AgNO3 (aq) AgCl (s) + KNO3 (aq)
K+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) AgCl(s) + K+(aq) + NO3-(aq)
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24. Concentration of Solutions
Qualitative Expressions of Concentrations
Dilute: a solution that contains a relatively small
amount of dissolved solute.
Example:
A 0.1 M HCl solution is dilute acid.
Concentrated: a solution that contains a relatively large
amount of dissolved solute.
Example:
A 12 M HCl solution is concentrated acid.
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25. Quantitative Expressions of Concentrations:
Concentration of Solutions
Quantitative Expressions of Concentrations:
Units
Symbol
Definition
Mass percent
% m/m
Part per million
ppm
Mass/Volume percent
% m/v
Volume percent
% v/v
Molarity M
Molality m
mass solute
mass solution
x 100
mass solute
mass solution
x 1,000,000
mass solute
mL solution
x 100
mL solute
mL solution
x 100
mol solute
L solution
mol solute
kg solvent
© 2014 John Wiley & Sons, Inc. All rights reserved.

26. Mass Percent Calculate the mass % of NaCl in a solution prepared
by dissolving 50.0 g of NaCl in g of H2O.
Knowns
50.0 g NaCl (solute mass)
150.0 g H2O (solvent mass)
200.0 g solution (solute + solvent mass)
Formula
mass solute
mass solution
mass % =
x 100
Calculate
50 g NaCl
200 g soln
mass % =
x 100 = 25% NaCl
© 2014 John Wiley & Sons, Inc. All rights reserved.

27. Mass Percent Practice What is the mass of Na2CO3 needed to make
350.0 g of a 12.3% aqueous solution?
Knowns
12.3% solution (mass %)
350.0 g solution (solute + solvent mass)
Solve for mass of solute (Na2CO3)
Formula
mass % x mass soln
100
mass solute
mass solution
mass % =
x 100
mass solute =
Calculate
12.3 x g
100
mass solute =
= 43.1 g Na2CO3
© 2014 John Wiley & Sons, Inc. All rights reserved.

28. Mass-Volume Percent Saline is a 0.9 m/v % NaCl solution. What mass of
sodium chloride is needed to make 50 mL of saline?
Knowns
50.0 mL solution (solution volume)
0.90 m/v% (mass/volume %)
Solve for mass of solute (NaCl)
Formula
g solute
mL solution
m/v % x mL soln
100
m/v % =
x 100
mass solute =
Calculate
50.0 x 0.90
100
mass solute =
= 0.45 g NaCl
© 2014 John Wiley & Sons, Inc. All rights reserved.

29. Mass-Volume Percent Practice
What volume of a 3.0% H2O2 solution
will contain 10.0 g of H2O2?
a mL soln
b mL soln
c L soln
d. 165 mL soln
Knowns
10.0 g H2O2 (desired solute mass)
3.0 m/v%
Solve for volume of solution
Formula
g solute
mL solution
mL solution =
g solute
m/v %
x 100
m/v % =
x 100
Calculate
10.0 g
3.0
mass solute =
x 100
= 330. mL sln
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30. Volume Percent What volume of soda that is 6.0 % by volume alcohol
contains mL of ethanol (CH3CH2OH)?
Knowns
20.0 mL ethanol (solute volume)
6.0 volume %
Solve for volume of solution
Formula
volume solute
volume solution
volume % =
x 100
volume solute
volume %
volume soln =
x 100
Calculate
200.0
6.0
volume soln =
x 100
= 3300 mL soda
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31. Volume Percent Practice
A solution is prepared by mixing 20.0 mL of propanol
with enough water to produce mL of solution.
What is the volume percent of propanol?
a %
b %
c %
d %
Knowns
20.0 mL propanol (solute volume)
400.0 mL solution (solution volume)
Solve for volume percent
Formula
volume solute
volume solution
volume % =
x 100
Calculate
20.0
400.0
volume % =
x 100
= 5.00% propanol
© 2014 John Wiley & Sons, Inc. All rights reserved.

32. A common unit for solution concentration dissolved in enough water to
Molarity
A common unit for solution concentration
due to convenience.
mol solute
L solution
molarity =
Example:
To prepare a 1.0 M KCl
solution, 1.0 mol of KCl is
dissolved in enough water to
make 1.0 L of solution.
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33. Molarity Practice Calculate the molarity of a solution prepared by
dissolving 9.35 g of KCl in enough water to
prepare a mL solution.
Knowns
9.35 g KCl (solute mass)
250.0 mL solution (solution volume)
mol solute
L solution
Formula
Solve for molarity
molarity =
Calculate
1 mol KCl
9.35 g KCl x
= mol KCl
g KCl
0.125 mol KCl
molarity =
= M KCl
0.250 L solution
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34. How many grams of KOH are required to prepare
Molarity Practice
How many grams of KOH are required to prepare
600.0 mL of a M KOH solution?
a g KOH
b g KOH
c x 104 g KOH
d g KOH
Knowns
600 mL (solution volume)
0.450 M (solution molarity)
Plan
Solve for moles, then grams using molarity and
molar mass as conversion factors
mol solute
L solution
Formula
molarity =
mol solute =
molarity x L soln
Calculate
moles solute = M KOH x L = mol KOH
56.11 g KOH
0.270 mol KOH x
= 15.1 g KOH
1 mol KOH
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35. Solution Stoichiometry
Similar to previous stoichiometry problems, but we can
now use molarity as an additional conversion factor.
How many mL of M Hg(NO3)2 are needed
to precipitate 2.50 g of KI?
Hg(NO3)2 (aq) + 2 KI (aq) KNO3 (aq) + HgI2 (s)
Plan
g KI mol KI mol Hg(NO3) mL soln
Calculate
1 mol KI
1 mol Hg(NO3)2
1000 mL soln
2.50 g KI × × ×
g KI
2 mol KI
0.175 mol Hg(NO3)2
= mL Hg(NO3)2
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36. Solution Stoichiometry Practice
How many grams of AgCl will form by adding enough
AgNO3 to react fully with mL of
0.400 M BaCl2 solution?
2 AgNO3 (aq) + BaCl2(aq) AgCl (s) + Ba(NO3)2 (aq)
a. 172 g AgCl
b g AgCl
c x 10-3 g AgCl
d g AgCl
Plan
Volume BaCl mol BaCl mol AgCl g AgCl
Calculate
0.400 mol BaCl2
2 mol AgCl
143.4 g AgCl
1500. mL × × ×
1000 mL
1 mol BaCl2
1 mol AgCl
= 172 g AgCl
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37. Dilution Dilution: Adding a solvent to a concentrated solution to
make the solution less concentrated (i.e. dilute).
When a solution is diluted, only the volume changes. The number of moles of solute remains constant.
moles before dilution = moles after dilution
Molarity1 x Volume1 = Molarity2 x Volume2
M1 × V1 = M2 × V2
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38. What volume of 12 M HCl is needed to make
Dilution Practice
What volume of 12 M HCl is needed to make
500.0 mL of a 0.10 M HCl?
Knowns
12 M HCl M1
0.10 M HCl M2
500.0 mL V2
Solving for:
volume of 12 M HCl V1
M1 × V1 = M2 × V2
Calculate
V2M2 M1
500 mL x 0.10 M
12 M
V1 = =
= 4.2 mL
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39. Dilution Practice
Calculate the molarity of a NaOH solution prepared by
mixing 100. mL of 0.20 M NaOH with 150 mL of H2O.
a. 2.0 M NaOH
b M NaOH
c M NaOH
d M NaOH
Knowns
0.20 M NaOH M1
100 mL sln V1
= 250 mL V2
Solving for:
molarity NaOH M2
M1 × V1 = M2 × V2
Calculate
M1V1 V2
0.20 M x 100 mL
250 mL
M2 = =
= M NaOH
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40. Colligative Properties of Solutions
Colligative Property: A solution property that depends
only on the number of solute particles not the
nature of the particles.
Common Colligative Properties:
1. Vapor Pressure Lowering
Solutions have lower vapor pressures than pure solvent.
2. Boiling Point Elevation
Solutions have higher boiling points than pure solvent.
3. Freezing Point Depression
Solutions have lower freezing points than pure solvent.
4. Osmosis and Osmotic Pressure
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41. Vapor Pressure Lowering
Dissolving a solute in a solvent lowers the
vapor pressure of the solvent.
As a result, the solvent’s boiling point is increased (a)
while the freezing point of the solvent is lowered (b).
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14-42

42. Molality Since colligative properties depend on the number of
particles in the solvent and not the identity,
a new concentration unit is used when
discussing colligative properties.
mol solute
kg solvent
molality =
Example:
What is the molality of a solution prepared by
dissolving 0.10 mol of starch in 0.50 kg of water?
0.10 mol
0.50 kg H2O
m =
= 0.20 m
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43. Molality Practice What is the molality (m) of a solution prepared by
dissolving 2.70 g of methanol (CH3OH) in 25.0 g of water?
a m
b m
c m
d m
1 mol CH3OH
2.70 g CH3OH ×
32.04 g CH3OH
= mol CH3OH
mol
kg H2O
m =
= 3.37 m
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44. Colligative Properties
Calculating the change in boiling/freezing point
of a solution:
molality
ΔT = m x K
Change in temperature
Freezing (Kf) or boiling (Kb) point constants
Freezing/Boiling Points and the Related K Constants
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45. Boiling Point Elevation
What is the boiling point of a solution prepared by
dissolving 0.10 mol of sugar in 0.50 kg of water?
(Boiling point of water is ºC and Kb = ºC/m)
0.10 mol
0.50 kg H2O
= 0.20 m
ΔTb = 0.20 m x ºC/m = 0.10 ºC
Boiling point is always elevated by added solute,
so the change in temperature is added to the
boiling point of pure water.
Tb = ºC ºC = ºC
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46. Boiling Point Elevation Practice
What is the boiling point of an aqueous solution
that is 4.00 m in solute?
Tb(pure water) = ºC and Kb = ºC/m
a ºC
b ºC
c ºC
d ºC
ΔTb = 4.00 m x ºC/m = 2.05 ºC
Boiling point is always elevated by added solute,
so the change in temperature is added to the
boiling point of pure water.
Tb = ºC ºC = ºC
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47. Freezing Point Depression
What is the freezing point of a solution prepared by
dissolving 0.10 mol of sugar in 0.50 kg of water?
(Freezing point of water is 0.0 ºC and Kf = 1.86 ºC/m)
0.10 mol
0.50 kg H2O
Molality =
= 0.20 m
ΔTf = 0.20 m x 1.86 ºC/m = 0.37 ºC
Freezing point is always depressed by added solute,
so the change in temperature is subtracted from the
freezing point of pure water.
Tb = 0.0 ºC – 0.37 ºC = ºC
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48. Freezing Point Depression Practice
What is the freezing point of 100. g of C2H6O2 dissolved in
200 g of H2O? Tf(pure water) = 0.0 ºC and Kf = 1.86 ºC/m)
1 mol C2H6O2
a ºC
b ºC
c ºC
d ºC
2.70 g C2H6O2 ×
= mol C2H6O2
62.07 g C2H6O2
1.61 mol
0.20 kg H2O
Molality =
= 8.05 m
ΔTf = 8.05 m x 1.86 ºC/m = 15.0 ºC
Freezing point is always depressed by added solute,
so the change in temperature is subtracted from the
freezing point of pure water.
Tf = 0.0 ºC – 15.0 ºC = ºC
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49. Osmosis Osmosis: diffusion of water from a dilute solution or
pure water, through a semipermeable membrane into a
solution of higher solute concentration.
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50. Osmotic Pressure
Osmotic Pressure: difference in the amount of pressure
necessary to apply to a solution to stop the flow of water
due to osmosis and the atmospheric pressure.
Demonstrating Osmotic Pressure
Water flows through the membrane into the more concentrated
sugar solution, causing the solution to rise in the tube.
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51. © 2014 John Wiley & Sons, Inc. All rights reserved.
Blood and Osmosis
Isotonic: same concentration of dissolved solute as a cell.
(0.9% saline)
Hypertonic: higher concentration of dissolved particles
relative to cellular levels. (1.6% saline)
Hypotonic: lower concentration of dissolved particles
relative to cellular levels. (0.2% saline)
Effect of Different Concentrations on Red Blood Cells
isotonic
hypertonic
hypotonic
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52. Learning Objectives 14.1 General Properties of Solutions
List the properties of a true solution
14.2 Solubility
Define solubility and the factors that affect it.
14.3 Rate of Dissolving Solids
Describe the factors that affect the rate
at which a solid dissolves.
© 2014 John Wiley & Sons, Inc. All rights reserved.

53. Learning Objectives 14.4 Concentrations of Solutions
Solve problems involving mass percent, volume percent,
molarity, and dilution.
14.5 Colligative Properties of Solutions
Use the concept of colligative properties to calculate
molality, freezing point, boiling point, freezing point
depression, boiling point elevation of various solutions.
14.6 Osmosis and Osmotic Pressure
Discuss osmosis and osmotic pressure and
their importance in biological systems.
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