Agenda:
Homework: Read from Chapter 15
to complete tables 3,4,7 & 8
 Warm-up: Explain the solvation process
 Review – terms, importance, “likes dissolve likes”
 Solubility Curves
 What can they tell us about solutions?
 Measuring concentration of solutions
 Peer review of lab reports
Warm-up: Solvation Process
 Explain what is happening when an ionic compound is
dissolved in water (called solvation or hydration)
 View the video - what are key steps that need?
 How does the water molecules interact with the ionic
compound?
 http://group.chem.iastate.edu/Greenbowe/sections/pro
jectfolder/flashfiles/thermochem/solutionSalt.html
Solution process
Homework:
Solutions: Chapter 15
 Read pp 453-457, 476-
479
 Complete tables in notes
3. Examples of solutions
4. Size of solute particles
7. Making solutions
Questions:
 States of Matter
Summary – Due on
Thursday
Why are solutions important?
6. Making Solutions: Molecules
(Covalent compounds)
 General Rule: “Likes dissolve likes”
Likes dissolve likes
Polar solvents will dissolve
polar solutes
Non-electrolytes – do not
separate into ions
Sugar
Ethanol
Water cannot dissolve non-polar
solutes
Insoluble
 Examples:
Organic solvents
 Non-polar solvent dissolving non-polar solutes
 Examples:
 To remove oily stains in dry-cleaning

Lipids will dissolve in hydrocarbon
 To clean oil based paints
 To manufacture plastics, man-made fibers, adhesives
7. Making Solutions: Factors that impact the
solubility of solids in liquids
 Energy is involved – 2 steps
 Endothermic to separate solute & solvent particles
 Exothermics – attraction between solvent & solute particles
Factors affecting solids
dissolving in liquids
Chemical components of solute and
solvent
Intermolecular forces
Temperature
Particle size
Agitation (amount of stirring)
General trend to increase
solubility
9. Measuring Concentration
 Solubility: number of grams of solute in 100 grams of
water at 20℃
 Solubility Curves
 Saturated solutions
 Unsaturated
 Supersaturated


http://www.youtube.com/watch?v=XSGvy2FPfCw
http://www.youtube.com/watch?v=1y3bKIOkcmk&feature=re
lated
Solubility & Temperature
Solute (g) per 100 g H₂O
 Reading curves
Line shows the
amount of solute in
a saturated solution
Super saturated
Unsaturated
Temperature
http://www.youtube.com/watch?v=XSGvy2FPfCw
http://www.youtube.com/watch?v=XSGvy2FPfCw
 The video begins with a few crystals of sodium acetate
placed on the lab bench. A supersaturated solution of
sodium acetate is poured over the crystals providing a seed
or crystallization. The salt begins to crystallize, forming a
large sodium acetate structure from the precipitation of the
ions out of solution. When the sodium acetate crystallizes,
the oppositely charged ions are brought closer together by
the crystal structure. Since formation of a crystal lattice
lowers potential energy by placing like charges close
together, the system releases the excess energy in the
crystallization process. Thus, the structure ends up being
warm to the touch from this excess energy
Examples of supersaturated
solutions
 A good example of supersaturation is provided by Na2S2O3, sodium
thiosulfate, whose solubility at 25°C is 50 g Na2S2O3 per 100 g H2O. If 70
g Na2S2O3 crystals is dissolved in 100 g hot H2O and the solution cooled
to room temperature, the extra 20 g Na2S2O3 usually does not
precipitate. The resulting solution is supersaturated; consequently it is
also unstable. It can be “seeded” by adding a crystal of Na2S2O3,
whereupon the excess salt suddenly crystallizes and heat is given off.
After the crystals have settled and the temperature has returned to
25°C, the solution above the crystals is a saturated solution—it contains
50 g Na2S2O3.
Interpreting
solubility curves
Y axis
X axis
Lines
If the amount needed is
more or less than 100 g
of H₂O
1 g H₂0 = 1 mL H₂0
Determine the amount of solute
required to make a saturated
solution
Amount, grams
KNO3at 70°C
NH4Cl at 90°C
NaCl at 100°C
KI at 20°C
NaNO3at 35°C
SO2 at 50°C
NH3 at 20°C
KClO3 at 65°C
KCl at 75°C
NH4Cl at 65°C
HCl at 10°C
NaNO3at 70°C
KNO3at 10°C
Determine the amount of solute
required to make a saturated
solution
Amount
in grams
KNO3at 70°C
NH4Cl at 90°C
NaCl at 100°C
KI at 20°C
NaNO3at 35°C
SO2 at 50°C
NH3 at 20°C
KClO3 at 65°C
KCl at 75°C
NH4Cl at 65°C
HCl at 10°C
NaNO3at 70°C
KNO3at 10°C
140g
72g
40g
145g
100g
5g
55g
20g
50g
60g
78g
135g
Gases
What additional
information is in
this curve?
Explain how gases
are different from
the other
compounds?
Type:
Size:
Temperature:
8. Making solutions: Factors that affect the
solubility of gases in liquids
Think about soda (carbonated drinks)
Factors affecting gases
Temperature
Pressure
General trends to increase
solubility
Practice:
Interpreting Solubility Curve
 Activity 5-5
 Define: solubility (quantitative )
 Saturated
 Precipitate
Solubility Rules determined under specified
concentration (often 0.1% molarity)
Except with
these ions
Category
Ions
Soluble cations
Group 1 ions
No exceptions
and ammonium, NH4+
NO3- and
Soluble anions
C2H3O2Usually soluble
anions
Cl-, Br-, and I-
SO42-
Usually
insoluble
anions
CO32-, PO43-, and OH-
No exceptions
Soluble:
Exceptions: Ag+
and Pb2+
Soluble
Exceptions:
Ba2+ and Pb2+
Examples
Na2CO3, LiOH, and
(NH4)2S are soluble.
Bi(NO3)3, and
Co(C2H3O2)2 are soluble.
CuCl2 is water soluble,
but AgCl is insoluble.
FeSO4 is water soluble,
but BaSO4 is insoluble.
CaCO3, Ca3(PO4)2, and
Insoluble:
Mn(OH)2 are insoluble in
Exceptions: group
water, but (NH4)2CO3,
1 elements and
Li3PO4, and CsOH are
NH4+
soluble.
9f. Measuring Concentrations
 Molarity (M) by definition =
 Molarity = Moles of solute
1 Liter of solution


Therefore: 2 molar solution = __________________
 Discovery Education video

Standard Deviants School Chemistry: solutions & dilutions
Molarity: moles of solute
liters of solution
Molarity problems
 What I s the molarity of a solution in which 58 gram of
NaCl are dissolved in 1.0 L of solution?
 What is the molarity of a solution in which 10.0 grama
of silver I nitrate is dissolved in 500 mL of solution?
 How many grams of potassium nitrate should be used
to prepare a 2.0 L of a 0.5 molar solution?
Steps needed in molarity
calculations
 Moles
For grams:
 Liters of solution
 If less than 1L ? mL = 1L
 Comparison __________ = moles of solute
1 liter of solution
Molarity practice
 To what volume should 5.0 g of KCl be diluted in order
to prepare a 0.25M solution?
 How many grams of copper II sulfate – penta hydrate
are needed to prepare 100 mL of a 0.10M solution?
9g. Dilution of concentrated
solutions:
V₁M₁= V₂M₂
 Molarity by dilution practice problems
 Most reagents are sold & sorted in concentrated
solutions
 How much concentrated 18M sulfuric acide is needed
to prepare 250mL of a 6.0M solution?
 How much concentrated 12M hydrochloric acid is
needed to prepare 100 mL of a 2.0 M solution?
 To what volume should 25 mL of 15 M nitric actid be
diluted to prepare a 3.0 M solution?
 To how much water should 50 mL of 12 M hydrochloric
acid be added to produce a 4.0 M solution?
 To how much water should 100 mL of 18M sulfuric acid be
added to prepare 1.5 M solution?
Colligative Properties
 The properties of the solution that depend on the
number of particles in solution, not the identity of the
solute.
 The solvent properties will be changed.
 NaCl
 CaCl₂
 AlCl₃
 Melting point depression; boiling point elevation;
 Vapor pressure lowering
Resources
 http://www.karentimber
 http://www.chemistryge
lake.com/solution.htm
 http://www.afn.org/~afn
02809/powerpointlist.htm
ek.com/chemistrypower
point/Student%20Ch%2
015%20Solutions.ppt
 http://college.cengage.co
m/chemistry/general/zu
mdahl/world_of_chem/1
e/instructors/ppt/figures
/viewindex.html
Solution process