If you are not part
of the SOLUTION,
you’re part of the
PRECIPITATE!
SOLUTIONS
Chapter 15
SOLUTION OBJECTIVES
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Differentiate between colloids, solutions, and suspensions
Define solvent and solute and be able to identify them in a
solution.
Describe the process of solvation, dissociation, and dissolving.
Be sure to study polarity of compounds to be able to predict
solubility.
Differentiate between saturated, unsaturated, and
supersaturated.
State and discuss the factors affecting the rate of solubility.
State and discuss the factors that affect solubility.
Be able to read a solubility curve graph.
Relate the enthalpy of solution to endothermic and
exothermic dissolving processes.
Differentiate between molarity and molality.
Solve problems involving molarity, molality, mole fraction
mass percent, volume percent, and making solutions.
THREE TYPES OF MIXTURES
SOLUTION
COLLOID
SUSPENSION
MIXTURES
DEMO
SOLUTION: Salt and water
COLLOID: Shaving gel
SUSPENSION: Cornstarch in water
TYNDALL EFFECT: particles
dispersed in mixture are big
enough to scatter light.
SOLUTIONS
A
solution is a
homogeneous mixture of
two or more substances
in a single physical state.
 SOLVENT:
the substance
doing the dissolving and
in the greater amount.
 SOLUTE: the substance
being dissolved and in
less abundant.
DEMO
 If
I combine 50mL beads and 50.0mL
sand, what will the final volume be?
 If
I combine 50.0mL water (green)
and 50.0mL alcohol, what will the final
volume be?
NINE BASIC TYPES OF SOLUTIONS
Liquid Solutions:
solid in liquid (salt water)
liquid in liquid (vinegar)
gas in liquid (carbonated drink)
Gaseous Solutions: All gaseous mixtures are
solutions
solid in gas (soot in air)
liquid in gas (humid air)
gas in gas (air)
NINE BASIC TYPES OF SOLUTIONS
Solid Solutions
Homogeneous mixtures of solids are usually
made from liquid solutions that have been
mixed and then solidified (frozen)
 solid in solid (alloys – brass, bronze)
 liquid in solid (dental fillings)
 gas in solid (charcoal gas mask)
SOLUTION EQUILIBRIUM
-
A solution is in
dynamic equilibrium
when the number of
solute particles
returning to the
crystal surface is
equal to the number
of solute particles
leaving the crystal
surface.
SOLUTION EQUILIBRIUM
A
SATURATED solution is a solution with the
maximum amount of solute dissolved at a
given temperature (Any more added goes to
the bottom). It has reached dynamic
equilibrium.
 An UNSATURATED solution has the ability to
dissolve more solute at a given temperature
 If a solution is SUPERSATURATED, then a hot
solution is saturated and cooled. An
unstable condition results because the
solution holds more solute that it normally
does at a given temperature.
LASERDISC
SOLUTIONS – Chapter 20
1.
2.
3.
What are ion-dipole attractions? During the process of
solvation, how do ion-dipole attractions compare to the
ionic bonds inside an ionic crystal?
Describe the processes that are balanced when
solubility equilibrium has been reached. If more solute is
added to a solution in equilibrium, will this always result in
more dissolved solute particles in the solution? Explain.
How does a solution become supersaturated? After
crystals precipitate out of a supersaturated solution,
would you still call it supersaturated? Explain.
THE DISSOLVING PRACTICE
 Dissolving
occurs when the solute is pulled apart
by the solvent.
 This takes place at the surface of the solute. The
solvent surrounds the solute.
 This process of surrounding the solute is called
SOLVATION.
 When the surrounding is done by water, this is
called HYDRATION, a particular type of
solvation.
 When ionic compounds separate into their ions
in a solvent, DISSOCIATION occurs.
DISSOCIATION OF NaCl
Solute-Solvent Combinations
1. Polar Solvent – Polar Solute:
The polar solvent is attracted to the
polar solute. The solvent gradually
surrounds the solute. The particles
attach themselves due to polar
attraction. Solvation occurs. Like
dissolves like. Ex. Salt and water
Solute-Solvent Combinations
2. Polar Solvent – Nonpolar Solute:
Polar solvent particles are attracted
to each other and not the solute.
Solvation does not occur and a
solution is unlikely. Ex. oil and water
DEMO: Marbles
Solute-Solvent Combinations
3. Nonpolar Solvent – Polar Solute:
Nonpolar solvent particles have
little attraction to the polar solute.
Solvation does not occur and a
solution is unlikely. Ex. Salt and oil
Solute-Solvent Combinations
4. Nonpolar solvent – nonpolar
solute:
Random motion of solute particles
causes them to leave the surface of
the solute and become evenly
dispersed in the nonpolar solvent.
Solvation occurs. Like dissolves like.
LASERDISC
Polar and Nonpolar Solvents – Chapter 30
1.
2.
3.
4.
Why are some liquids immiscible? Explain in terms of
intermolecular forces?
How would you predict whether carbon tetrachloride is a
polar or nonpolar solvent? What evidence have you
observed that supports your prediction?
Based on chemical formula alone, can you tell whether
iodine, I2, is a polar or nonpolar solute? What evidence
have you observed that supports your prediction?
Based on the chemical formula alone, can you tell
whether copper (II) chloride, CuCl2, is a polar or nonpolar
solute? What evidence have you observed that supports
your prediction?
DO NOW
Pick
up handout – due Tuesday
Turn in Superaturation lab
Book projects due today: William
B., Marwa J., Quinn R., Devon L.,
Andrew C., Madison B., Peter D.,
Seth T., Conner R., Kat M., Tatum
N., Sam W., Dan N., Claire W., and
Jamal R.
SOLUBILITY
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Solubility is how much solute can dissolve in a
given amount of solvent.
It is measured in g/L or mol/L. It is usually the
grams of solute per 100g of solvent.
A CONCENTRATED solution is said to have a
high ratio of solute to solvent.
A DILUTE solution is the opposite of this.
For solids: solubility usually increases with
increased temperature
For gases: solubility usually decreases with
increased temperature
SOLUBILITY
How do you
determine saturated,
unsaturated, and
supersaturated for a
given substance?
FACTORS AFFECTING SOLUBILITY
1.
Nature of the solvent and solute “like dissolves like”. This means that
polar solvent dissolves a polar
solute and nonpolar solvent
dissolves nonpolar solute. But
polar does not dissolve nonpolar.
In other words – will it even dissolve?
FACTORS AFFECTING SOLUBILITY
2.
Temperature increase the
temperature
and solubility
increases
(except gases).
FACTORS AFFECTING SOLUBILITY
3. Pressure - increase the pressure and
you increase solubility (only with
gases).
Gases ONLY
4. How much is already dissolved –
saturated versus unsaturated versus
supersaturated…
Factors Affecting RATE of Solubility
(How fast something will dissolve)
 Agitation
(shaking, stirring)
 Increased temperature (except gases)
 Smaller Particle size
Each allows more solvent to come in contact
with the solute faster.
HEAT OF SOLUTION
 Energy
is released and absorbed as substances
dissolve.
 ENDOTHERMIC: The solute particles must
separate and the solvent particles must
separate to make room for the solute.
 EXOTHERMIC: When the solute and the solvent
mix, the particles are attracted to each other.
 The overall change in energy is the heat of
solution.
 Most solutions are endothermic.
CONCENTRATION
Concentration is NOT dependent upon the sample
size and it can be measured.
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Grams/100.0 grams
measures solubility
Parts per million (ppm)
measures small concentrations
Parts per billion (ppb)
measures pollutants
Molarity
used in lab chemistry
Molality
used for special calculations
We will learn to calculate molarity, molality, mass
percent, volume percent, and mole fraction
MOLARITY, M
This is the ratio between the moles of dissolved
substance and the volume of the solution
expressed in liters.
Molarity, M =
moles of solute
volume of solution in liters
A one-molar (1M) solution of HCl contains one
mole of HCl in one liter of water. (Which means
it contains 36.46g of HCl in 1 liter of water.)
SAMPLE PROBLEM
Sandy dissolves 45.0 g of NaCl in 2.5 liters
of solution. What is the concentration in
molarity of NaCl?
Mass = 45.0 g
V = 2.5 L
Molar Mass = 58.44 g/mol
Molarity = 45.0 g NaCl
2.5 L
= 0.31M
1 mol NaCl
58.44 g NaCl
PRACTICE
What is the molarity of 58.5g of NaCl
dissolved in 2.0L of solution?
MOLALITY, m
This is concentration expressed in terms of
moles of solute per kilogram of solvent.
Volume is not a factor.
Molality, m = moles solute
Kg solvent
A 1.0 molal aqueous sugar soln:
1 mole sugar
1 kg water
SAMPLE PROBLEM
Calculate the molality of 98.0g RbBr in
0.824 Kg water.
m =
98.0 g RbBr
0.824Kg H2O
= 0.719m
1 mol RbBr
165.38g RbBr
PRACTICE
Calculate the molality of 85.2g SnBr2
in 140.0g water.
MASS PERCENT
Scientists frequently express the
concentration of solutions in mass
percent.
Mass % =
g of solute
x 100
g of solute + g of solvent
A 5% solution of NaOH contains 5g NaOH in
each 100g of solution (95g solvent and 5g
of solute).
VOLUME PERCENT
Consumer products frequently express their
concentration of solutions in volume
percent.
Volume % =
L of solute
x 100
L of solute + L of solvent
A 5% solution of sodium hydroxide contains
.05L NaOH in each 1.0L of solution (0.95L
solvent and 0.05L of solute).
MOLE FRACTION
The concentration of solution can
also be expressed in mole fractions.
Mole Fraction =
mole of solute
mole of solute + mole of solvent
PREPARING SOLUTIONS
A 3M solution of HCl is not bought but
made from 12M stock solutions.
In addition, a 1.0M solution of NaOH is
made from a calculated amount of
solid NaOH added to water.
It is important to know how to make
different concentration of solutions.
PREPARING SOLUTIONS
Prepare 1.0 liter of a 1.0M aqueous solution
of NaOH
M = mol
1.0M = mol
L
1.0L
mol = (1.0M)(1.0L) = 1.0mol
1.0mol NaOH 40.00g NaOH = 40.00g
1 mol NaOH
So, you will put 40.00g NaOH in a flask, add
1.0L water, and mix well.
Preparing Solutions
Prepare 1.0L of a 6.0M aqueous
solution of KCl. The molar mass of
KCl is 74.55g/mol.
DILUTING SOLUTIONS
To dilute a solution, you can form a
ratio between molarity and volume.
M1V1 = M2V2
DILUTING SOLUTIONS
Diluting a 12.0M solution of HCl to 6.0M HCl
What volume would you use to make 0.500L
of 6.0M HCl solution?
V1 = M2V2 = (6.0M HCl)(0.500L HCl)
M1
(12.0M)
= 0.250L HCl
Thus you would need 0.250L HCl and 0.250L
water to make the solution.
PRACTICE
Preparing a 0.1M solution of HCl
What volume would you use of 12.0M
HCl to make 1.0L of 0.1M HCl?
ANSWER
Preparing a 0.1M solution of HCl
M1 = 12.0M HCl
V1 = ?
M2 = 0.1M HCl
V2 = 1.0L
V1 = M2V2
M1
V1 = (0.1MHCl)(1.0L)
(12.0M HCl)
= 0.0083L
M1V1 = M2V2
Add 8.3mL 12.0M HCl to
991.7mL water
COLLIGATIVE PROPERTIES
 “colligative”
– depending upon the collection
 These
are properties that depend on the
concentration (the number of) of the solute
particles.
Three Factors:
 Vapor pressure reduction
 Boiling Point Elevation
 Freezing Point Depression
LASERDISC
Colligative Properties (chapter 32)
1.
Support one of the following hypotheses about why it
helps to add salt to the water before cooking the
pasta: a) The salt brings the water to a boil sooner; or
b) The salt brings the water to a boil at a higher
temperature.
2.
Which antifreeze solution boils at the highest
temperature? Which solution would you want in your
car on a hot summer day? Explain your reasoning.
3.
On cold winter days, unprotected radiator water can
freeze and expand, which can ruin an engine by
cracking the engine block. Which antifreeze mixture
would you want in your car on a cold winter day?
Vapor Pressure: A Review
VAPOR PRESSURE REDUCTION
VAPOR PRESSURE REDUCTION
VAPOR PRESSURE REDUCTION
 The
vapor pressure of a solvent
containing a non-volatile solute is
LOWER than the vapor pressure of the
pure solvent.
 WHY:
The solute takes up space at the
surface of the liquid and prevents some
solvent molecules from leaving the
liquid.
RAOULT’S LAW
 The
vapor pressure of a solution of a
non-volatile solute is equal to the
vapor pressure of the pure solvent
at that temperature multiplied by its
mole fraction.
RAOULT’S LAW
Po is the vapor pressure of the pure
solvent at a particular temperature.
xsolv is the mole fraction of the solvent.
So if the solution is 20% solute and 80%
solvent, the mole fraction for the solvent
is 0.8.
So the Vpsoln = 0.8 vpsolvent
RAOULT’S LAW
Because changes in state depend
upon vapor pressure, the presence of
a solute also affects the freezing
point and the boiling point of a
solvent.
**It depends on the number and not
the identity of the solute particles
BOILING POINT REVIEW
 The
boiling point of a
substance is the
temperature at which
the vapor pressure of a
liquid is equal to the
external pressure on its
surface (usually
atmospheric pressure).
BOILING POINT ELEVATION
 An
addition of solute lowers that vapor
pressure, therefore a higher temperature
is necessary to get the vapor pressure up
to the external (atmospheric) pressure so
that the solution will boil.
 Tb is the difference between the boiling
point of the solution and the boiling point
of the pure solvent. It is directly
proportional to the number of solute
particles per mole of solvent particles.
BOILING POINT ELEVATION
Dissociation factor
molality
Tb = (Kb)(df)(m)
constant (molal BP elevation constant)
The dissociation factor, df, is how many
particles the solute breaks up into.
For a nonelectrolyte, the df is 1
Water’s “Kb” value is 0.515C/m
FREEZING POINT: A REVIEW
 The
freezing point of a substance is
the temperature at which the vapor
pressure of the solid and liquid
phases are the same.
FREEZING POINT DEPRESSION
Club Soda demo
Animation for Freezing Point depression
FREEZING POINT DEPRESSION
 When
a dissolved solute lowers the
freezing point of its solution, you have
freezing point depression. Tf is directly
proportional to the molality of the solution.
 Tfp = (Kfp)(df) (m)
constant (molal fp depression constant)
Water’s “Kfp” value is 1.853C/m
CHEMICALS USED TO MELT ICE
Formula
Lowest Temp
Pros
Cons
(NH4)2SO4
-7.0°C
fertilizer
Damages concrete
CaCl2
-29.0°C
Melts ice faster
than NaCl
Attracts moisture;
surfaces can be
slippery
MgCl2
-15.0°C
Melts ice faster
than NaCl
Attracts moisture
CH3COOK
-9.0°C
biodegradable
corrosive
KCl
-7.0°C
fertilizer
Damages concrete
NaCl
-9.0°C
Keeps sidewalks
dry
Corrosive; damages
concrete and
vegetation
CMA
-9.0°C
Safest for
concrete and
vegetation
Works better at
preventing re-icing
CALCULATIONS: NON-IONIC
Determine the freezing and boiling point
for 85.0g of sugar (C12H22011) in 392g
water.
Need these things:
1. Molality, m
2. number of particles (should be one)
3. Change in temperature,
CALCULATIONS: NON-IONIC
Molality
392 g H2O
1.
1 kg
1000 g
m = 85.0 g C12H22011
0.392 Kg H2O
= 0.633m
2.
= 0.392Kg
1 mol C12H22011
342.34 g C12H22011
df = 1 (non-ionic solute)
CALCULATIONS: NON-IONIC
Change in Freezing point temperature:
Tfp = (Kfp)(df) (m) = 1.853C 1 0.633 m
m
= 1.17C
New Freezing Point:
0.00C – 1.17C = -1.17C
Freezing point of pure water
CALCULATIONS: NON-IONIC
Change in boiling point temperature:
Tbp = (Kbp)(df) (m) = 0.515C 1 0.633 m
m
= 0.326C
New Boiling Point:
100.000C + 0.326C =
100.326C
Boiling point of pure water
PRACTICE
What is the freezing point for a solution of 210.0g
of glycerol dissolved in 350.0g of water? (The
molecular mass of glycerol is 92.11 g/mol)
 FP = ?
 Solute = 210.0g glycerol
 Solvent = 350.0g H2O
PRACTICE
What is the freezing point for a solution of 210.0g
of glycerol dissolved in 350.0g of water? (The
molecular mass of glycerol is 92.11 g/mol)
1. m = 210.0g gly 1 mole gly 1000g
350.0g H2O 92.11 g gly 1 kg
= 6.51395 = 6.514m
2. Df = 1
3. Tf = (Kf)(df)(m) = (1.853ºC/m)(1)(6.514m)
= 12.070 = 12.07ºC
New FP = 0.000ºC – 12.07ºC = -12.07ºC
CALCULATIONS: IONIC
Determine the freezing and boiling
point for 21.6 g NiSO4 in 100.0 g water.
Need these things:
1. molality
2. number of ions
3. change in temperature
CALCULATIONS: IONIC
Calculate molality:
100.0g H2O
1 kg
1000 g
1.
m=
21.6 g NiSO4
0.1000 Kg H2O
= 1.396 = 1.40 m
= 0.1000 Kg
1 mol NiSO4
154.76 g NiSO4
CALCULATIONS: IONIC
2.
Determine the number of ions:
Substance is NiSO4
There are two ions, Ni+2 and SO4-2,
so the df is 2
CALCULATIONS: IONIC
3.
Calculate the change in temperature
and the new freezing point.
Tfp = 1.853C 2 1.40 m
m
0.00C – 5.19C =
-5.19C
= 5.19C
CALCULATIONS: IONIC
Tbp = 0.515C 2 1.40 m
m
100.00C + 1.44C =
= 1.44C
101.44C
CALCULATIONS: IONIC
What is the boiling point of a solution
containing 34.3 g of Mg(NO3)2 dissolved in 0.107
kg of water? (The formula mass of magnesium
nitrate is 148.32 g/mol.)
 Bp
=?
 Solute 34.3g Mg(NO3)2
 Solvent 0.107kg
CALCULATIONS: IONIC
1. m = 34.3g Mg(NO3)2 1 mole Mg(NO3)2
0.107kg
148.32 g Mg(NO3)2
= 2.16127796 = 2.16m
2. Df = 3
Mg+2
NO3-1
NO3-1
3. Tb = (Kb)(df)(m) = (0.515ºC/m)(3)(2.16m)
= 3.3372 = 3.34ºC
New BP = 100.0000ºC + 3.34ºC = 103.34ºC
Do Now
 Pick
up handout – due Monday as a
homework check
 Ice Cream Supplies due NEXT Tuesday
 Book Projects due Thursday: Hayden R.,
Matt H., Bob Wiley, Sean R., Sam K.,
Cameron H., Joe F., Marshall C., Josh
M., Kainath M., Grace M., Sarah., Adiba
K., Alexander G., and Stilian I.
RECAP
A
solution reduces the vapor pressure of
the pure solvent.
 Vapor pressure reduction results in
freezing point depression (lowering) and
boiling point elevation.
 To solve these problems, you need:



Molality
Dissociation factor (Number of ions)
Molal BP/FP constant (given to you)
RECAP
 Calculating
freezing point change:
 Tfp = (Kfp)(df)(m)
 Calculating
boiling point change:
Tb = (Kb)(df)(m)
Final step is to subtract from true freezing
point or add to true boiling point
DETERMINING MOLAR MASS
 Colligative
properties provide a useful
means to experimentally determine the
molar mass (molecular mass in one mole)
of an unknown substance.
 Steps:
1. Solve for molality.
2. Solve for moles of solute.
3. Solve for molar mass
This is basically going backwards
DETERMINING MOLAR MASS
A 10.0 g sample of an unknown organic
compound is dissolved in 0.100 kg water.
The boiling point of the solution is elevated
to 0.433oC above the normal boiling point
of water. What is the molar mass?
solute = 10.0g
Solvent = 0.100kg water
Tbp = 0.433oC
df = 1
DETERMINING MOLAR MASS
Step One: determine molality
Tbp = Kbp df m
m = Tbp
Kbp
= 0.433oC = 0.841 m (mol/kg)
0.515oC/m
Assume df = 1 because it is an unknown
ORGANIC compound (not ionic)
DETERMINING MOLAR MASS
Step Two: Detemine moles of solute
m = mol solute
Kg solvent
mol solute = m x kg solvent
= 0.841 mol/kg x 0.100 kg
= 0.0841 mol
DETERMINING MOLAR MASS
Step Three: Determine Molar Mass
mol solute =
mass solute
molar mass solute
molar mass = mass solute = 10.0 g
mol solute
0.0841 mol
= 118.90606 = 119 g/mol
DETERMINING MOLAR MASS
PRACTICE:
A solution of 3.39 g of an unknown in 10.00 g of water has a
freezing point of –7.31oC. The solution is a nonelectrolyte. What
is the molar mass of the compound? (The molal freezing point
constant of water is 1.853C/m.)
Mass solute = 3.39g
Mass solvent = 10.00g
FP = –7.31oC
ΔTfp = 7.31oC
df = 1
DETERMINING MOLAR MASS
PRACTICE:
A solution of 3.39 g of an unknown in 10.00 g of water has a
freezing point of –7.31oC. The solution is a nonelectrolyte. What
is the molar mass of the compound? (The molal freezing point
constant of water is 1.853C/m.)
Step One: determine molality
Tfp = Kfp df m
m=
Tfp =
7.31oC
= 3.94m
Kbp df
(1.853oC/m)(1)
DETERMINING MOLAR MASS
PRACTICE:
A solution of 3.39 g of an unknown in 10.00 g of water has a
freezing point of –7.31oC. The solution is a nonelectrolyte. What
is the molar mass of the compound? (The molal freezing point
constant of water is 1.853C/m.)
Step Two: Determine moles of solute
m = mol solute
Kg solvent
mol solute = m x kg solvent
= 3.44 m x 0.01000 kg
= 0.0394 mol
DETERMINING MOLAR MASS
PRACTICE:
A solution of 3.39 g of an unknown in 10.00 g of water has a
freezing point of –7.31oC. The solution is a nonelectrolyte. What is
the molar mass of the compound? (The molal freezing point
constant of water is 1.853C/m.)
Step Three: Determine Molar Mass
mol solute =
mass solute
molar mass solute
molar mass =
mass solute = 3.39g
mol solute
0.0394 mol
= 86.040609 = 86.0g/mol
SEIZURE
-
-
-
-
Hiorns began by reinforcing the walls and
ceiling, and tanking the flat
(apartment)with plastic sheeting.
Then 70-80,000 liters of hot copper sulfate
solution was pumped in through a hole in
the ceiling from the flat above.
Weeks went by, until the temperature of
the solution dropped, and the crystals
began to precipitate.
Finally, any remaining liquid was pumped
back out, to be recycled by the chemical
industry.