As you come in,
• The Materials:
– Paper and pencil
– Pick up calendar, assessment plan, and practice packet
• The Plan:
– Self-Assessment: Cumulative Quiz 7
– Visual/Auditory Activity: Solubility notes, including
polarity, solubility rules, exceptions to rules, and solubility
graphs
– Collaborative Partners: Interpreting Graphics WS –
Solubility Graphs
– Individual Practice: Complete Interpreting Graphics WS
– Practice Packet: “Reading Solubility Graphs” and
“Worksheet: Solubility Graphs”
• The Assessment:
Musical Theme:
– Solubility Graph Quiz- Wednesday
Motown Monday
– Concentration Quiz - Friday
Georgia Performance Standard SC7:
Students will
characterize the properties that describe solutions and the
nature of acids and bases.
The Water Molecule
•Triatomic
•Covalent
•Polar bonds & polar
overall
•Bent at 105° angle
Intermolecular Forces
 Polar molecules are
attracted to one
another by dipole
forces
 Water is attracted to
other water
molecules by a
special dipole force, a
hydrogen bond
Water Solutions
 “Chemically pure water never exists in nature
because water dissolves so many substances.” textbook
 Universal solvent
 Aqueous solutions
Water Solutions
 Ionic compounds
most readily dissolve
in water due to
extreme polarity
 Polar covalent
compounds also
dissolve in water
 Nonpolar compounds
don’t
Solvation of Ionic Compounds
A Few Exceptions...
 Remember the solubility
rules...
 In some ionic
compounds, the ions are
so attracted to each other
that they won’t break
apart and dissolve.
 These are INSOLUBLE
ionic compounds.
Solvation of Covalent Compounds
 Covalent compounds do NOT break apart in water
when dissolving.
 Solvation of covalent compounds means that each
solute molecule is surrounded by water molecules.
Conductors
 In general, aqueous
solutions of ionic
compounds are
electrolytes.
 Generally, aqueous
solutions of covalent
compounds are
nonelectrolytes.
As you come in,
• The Materials:
– Paper, calculator, practice packet and pencil
• The Plan:
– Opening Activity: Go over Unit 8 Test
– Self-Assessment: Review Solubility Graph, Answer
questions in packet and on Interpreting Graphics WS
– Visual/Auditory Activity:
– Concentrations notes, including molarity and molality
– Individual Practice: Molarity & Molality Practice
– 15-2 Practice Problems WS 1-5, 7, 9, 13, 17
• The Assessment:
– Solubility Graph Quiz- Wednesday
– Concentration Quiz - Friday
Musical Theme:
2-Step Tuesday
Solution Vocabulary
Solute: Dissolves in the solvent
Soluble: Able to be dissolved in the solvent
(applicable to any states of matter)
Insoluble: Unable to be dissolved in the solvent
(applicable to any states of matter)
Miscible: Able to be dissolved in the solvent
(applicable to liquid/liquid solutions)
Immiscible: Unable to be dissolved in the solvent
(applicable to liquid/liquid solutions)
Solubility Vocab
 Unsaturated: less than maximum amount of
solute is dissolved in the solvent
 Saturated: maximum amount of solute is
dissolved in the solvent
 Supersaturated: special conditions have been
created to dissolve more than maximum amount
of solute in the solvent
Determines IF Solute Will
Dissolve...
 The nature of the solvent and solute governs
whether a solute will solvate in a particular
solvent. Specifically, the nature of the
intramolecular bond. Polar molecules will
solvate with polar molecules. Nonpolar
molecules will solvate with nonpolar
molecules. BUT, polar and nonpolar will not
form solutions together.
Determines Speed of Dissolving
BRING SOLUTE IN
CONTACT WITH SOLVENT
1.
Agitation: Create more collisions mechanically
2. Temperature: More kinetic energy creates more
collisions
3.
Surface Area: Dissolving process is a surface
phenomenon, the more surface of the solute that
is exposed the faster the solvation
Determines How Much Will
Dissolve
SOLUBILITY: HOW MUCH WILL DISSOLVE
1.
Temperature: solubility of solid solute
increases as the temp. increases; solubility of
gaseous solute decreases as temp increases
ex. Hot water bubbles, thermal pollution
2. Pressure: solubility of gaseous solute
increases as the pressure increases
-Henry’s Law ex. Soft drinks
Temperature & Solubility
 Higher the
temperature, the
more solid will
dissolve in a liquid
 Higher the
temperature, the less
gas will dissolve in a
liquid
Henry’s Law:
Pressure & Solubility
 At a given
temperature, the
solubility of a gas is
proportional to the
pressure of the gas
above the liquid.
 Page 506
The higher the pressure, the more carbon dioxide will
dissolve in the syrup giving a less “flat” taste.
Solubility Graph
Concentration of Solutions
The concentration of a solution is a measure of how
much solute is dissolved in a specific amount of
solvent or solution.
Molarity: most common units of solution
concentration; # of moles solute dissolved in one liter
of the solution
Molarity: moles of solute
liter of solution
If given grams, remember to
change it to moles.
Example 15.3 (page 531)
Calculate the molarity of a solution prepared
by dissolving 11.5 g of solid NaOH in enough
water to make 1.50L of solution.
Example 15.4 (page 532)
Calculate the molarity of a solution prepared
by dissolving 1.56 g of gaseous HCl into
enough water to make 26.8 mL of solution.
The most common concentration
term in chemistry is Molarity (M),
but chemists also report
concentration in molality (m)
sometimes.
molality: moles of solute
kg of solvent
If given grams, remember to
change it to moles.
The front of the last page in the calculations packet
refers to Molality. Complete 1 (a), 2 (a), 3 (b), and 4
(b).
NOTE: Questions 3 and 4 will require you to
manipulate the equation. Question 4 might even
require that you change the units of your answer at
the end of the calculation.
As you come in,
• The Materials:
– Paper, calculator, practice packet and pencil
– Get a remote control
• The Plan:
– Opening Activity: Go over Solubility Graphs
– Assessment: Solubility Graph Quiz
– Kinesthetic Activity: Preparing a Solution Lab
– Individual Practice: Molarity & Molality Practice
– 15-2 Practice Problems WS 6, 8, 10-12, 14-16
• The Assessment:
– Concentration (Molarity & Molality) Quiz – Friday
(25pts)
Musical Theme:
Relaxing
Wednesday
Preparing a Solution in the Lab
 You’ve learned about concentration (Molarity and
molality). Therefore, you should be able to create your
own solutions for use in the lab from now on.
 Example:





0.5 M HCl reacts with Mg(s)
If I gave you HCl powder, what would you do to make the
solution?
Watch these kids, and assess your plan.
Choose a lab group and station. Draw an assignment from the
cup, and make the solution. Write your steps (including
materials) as you go.
ALSO: Solve 15-2 Practice Problems WS 6, 8, 10-12, 14-16.
As you come in,
• The Materials:
– Paper, calculator, practice packet and pencil
• The Plan:
– Opening Activity: Molarity/molality Practice
– Visual/Auditory Activity: Dilutions and
Stoichiometry
– Individual Practice: Solutions Stoichiometry in
practice packet
• The Assessment:
– Concentration (Molarity & Molality) Quiz – Friday
(25pts)
Musical Theme:
Disney Day
Diluting Solutions
If you already have a solution molarity, but you want a
different molarity:
Example: Your lab asks you to use 250 mL of 0.25M
HCl, but you only find a jug of 6M HCl in the
stockroom.
USE THIS EQUATION:
M1V1 = M2V2
Example 15.8 (page 539)
What volume of 16 M sulfuric acid must be used
to prepare 1.5 L of a 0.10 M H2SO4 solution?
*Be sure to solve the Dilution Worksheet in your
practice packet!
Using Molarity as a Conversion Factor
 You have learned to calculate molarity using the
equation:
 Molarity = moles of solute / liter of solution
 Did you realize that the calculation is simply a ratio of
solute to solution?
 Ratios (ie mole ratios, energy to mole ratios) can be
used as conversion factors in stoichiometry.
 THEREFORE, molarity can be written into our
dimensional analysis charts to solve stoichiometry
problems.
Solutions Stoichiometry
 What volume of 1.5 M HCl is needed to react with 21.5
grams of NaOH?
 What is the molarity of a solution of H2S if 48.5mL are
required to titrate 35.6mL of 0.35M Fe(OH)3 solution?
 A white precipitate forms when 200 mL of 0.200M
K3PO4 solution is mixed with 300 mL of 0.250 M
CaCl2 solution. What mass of precipitate will form?
 BE SURE TO SOLVE THE PRACTICE PROBLEMS IN
YOUR PACKET!
As you come in,
 The Materials: Remote control, paper, pencil,
calculator, practice packet
 The Plan:
 Opening Activity: Concentration Quiz
 Self-Assessment Activity: Go over Stoichiometry Exit
Question
 Visual/Auditory Activity: Colligative Properties notes
Individual Practice: Solving 15-4 colligative properties
calculations in packet
 The Assessment:
 Colligative Properties Quiz (15 pts) - Tuesday
Musical Theme: Greatest Hits Friday
Physical properties of solution are different
from the physical properties of the solvent.
Some properties are different simply
because there are “foreign” particles
(solute) in the solvent. Colligative properties
of solutions depend only on the number of
solute particles.
Number of Particles of Solute
 Think back to what you’ve learned about ionic and
covalent compounds dissolving in water.
 Ionic compounds DISSOCIATE as they dissolve.
 Covalent compounds do NOT dissociate as they
dissolve.
 Apply this idea.
 How many solute particles will 1 “piece” of NaCl add
to a solvent?

Answer: 2 (Na+ and Cl-)
Applying the Concept Further
 How many solute particles will 1 piece of magnesium
phosphate add to a solvent?
 Mg3(PO4)2 breaks into 3 Mg2+ ions and 2 PO43- ions.
Answer: 5 solute particles
 How many solute particles will 1 piece of sugar add to a
solvent?
 Sugar is covalent. Answer: 1 solute particle
 Finally, which of the two above will affect a colligative
property the most?
Boiling point is changed when
solute is added to the solvent.
• Boiling point elevates when a solute is
added to a solvent.
• The solution requires more energy to
reach boiling.
• Example: Salt water will not boil at
100°C. It will boil at a HIGHER temp.
As you come in,
• The Materials:
– Paper and pencil
• The Plan:
– Visual/Auditory Activity: Review colligative
properties
– Collaborative Partners: Colligative properties
practice
• The Assessment:
– Colligative Properties Quiz: Tuesday
– Cumulative Quiz: Thursday
– Unit Test: Friday
Musical Theme:
Motown Monday
Calculating the NEW Boiling Pt.
∆Tb = Kb m i
∆Tb = change in boiling pt.
Kb = boiling point constant for
the solvent (will be given)
m = molality
i = number of ions present in the
solute (USE ONLY WITH IONIC
SOLUTES!)
Example of Boiling Pt. Calculation
 What is the boiling point when 15.0g NaCl is
dissolved into 200 mL of water? (Kb of water is 0.52
°C/m)
 ∆Tb = Kb m i



You are solving for ∆Tb , and you have the Kb to use.
m =You’ll need to calculate the molality from the info in
the question. (change 15.0 g of NaCl to moles and 200 mL
to kg and plug in)
i = Finally, is the solute ionic? YES…NaCl is ionic and will
give TWO ions when dissolved.
Kb of water = 0.52°C/m
CHALLENGING:
How many grams of NaCl would need to be
added to the water to change the boiling
temperature of 200 mL to 110°C?
10C = m(0.52)(2)
m = 9.62
9.62m = mole/0.2kg
mole = 1.92 mole NaCl
112.2 grams NaCl
Freezing point is changed when
solute is added to the solvent.
• Freezing point depresses when a solute
is added to a solvent.
• The solution requires a lower temp to
reach freezing.
• Example: Salt water will not freeze at
0°C. It will freeze at a LOWER temp.
Calculating a NEW Freezing Pt.
∆Tf = Kf m i
∆Tf = change in freezing pt.
Kf = freezing point constant for
the solvent (will be given)
m = molality
i = number of ions present in
solute (USE ONLY WITH IONIC
SOLUTES)
Example of Freezing Pt. Calculation
 What is the freezing point when 15.0g NaCl is
dissolved into 200 mL of water?
 (Kf of water is 1.86 °C/m)
 ∆Tf = Kf m i



You are solving for ∆Tf , and you have the Kf to use.
m =You’ll need to calculate the molality from the info in
the question. (change 15.0 g of NaCl to moles and 200 mL
to kg and plug in)
i = Finally, is the solute ionic? YES…NaCl is ionic and will
give TWO ions when dissolved.
Kf of water = 1.86°C/m
CHALLENGING:
How many grams of NaCl would need to be
added to the water to change the freezing
temperature of 200 mL to -8°C?
8C = m(1.86C/m)(2)
molality = 2.15
2.15 = mole/0.2kg
mole = 0.43 mole NaCl
25.14 grams NaCl
Practice Practice Practice!
 The 15-4 Practice Problems in your practice packet are
boiling point/freezing point calculations.
 The key is posted at the front of the classroom.
 Colligative Properties Quiz - Tuesday
 PLEASE SOLVE MANY OF THESE PROBLEMS! You’ll
see several on Friday’s test.
As you come in,
 The Materials: Paper, pencil, calculator, practice
packet
 The Plan:
 Opening Activity: “We Solute You”
 Collaborative Practice: Solving “Colligative properties
problems” in packet
 Pre-Lab Activity: Prepare for Ice Cream Lab
 The Assessment:
 Colligative Properties Quiz (15 pts) - Tuesday
Musical Theme: Motown Monday
About.com
Chemistry
 Ice has to absorb energy in order to melt, changing the phase of water
from a solid to a liquid. When you use ice to cool the ingredients for ice
cream, the energy is absorbed from the ingredients and from the
outside environment (like your hands, if you are holding the baggie of
ice!). When you add salt to the ice, it lowers the freezing point of the
ice, so even more energy has to be absorbed from the environment in
order for the ice to melt. This makes the ice colder than it was before,
which is how your ice cream freezes. Ideally, you would make your ice
cream using 'ice cream salt', which is just salt sold as large crystals
instead of the small crystals you see in table salt. The larger crystals
take more time to dissolve in the water around the ice, which allows for
even cooling of the ice cream.