Lecture Notes
Chem 150 - K. Marr
Chapter 13
Properties of Solutions
Silberberg 3 ed
Properties of Solutions
13.1 Types of Solutions: IMF’s and Predicting
Solubility
13.2 Energy Changes in the Solution Process
13.3 Solubility as an Equilibrium Process
13.4 Quantitative Ways of Expressing Concentration
13.5 Colligative Properties of Solutions
Will not cover:
13.6 The Structure and Properties of Colloids
Definitions
1.
Solution
»
2.
homogeneous mixture with only one phase
present
Mixture
»
»
»
2 or more substances physically mixed together
Composition is variable
Properties of components are retained
Formation of Solutions
1.
Driving Force:
• Tendency toward Randomness (T1)
• Nature favors processes that result in an increase
in entropy (more randomness or less order)
– Explains why solutions of gases always form .
2.
Why don’t solutions always form with solids
and liquids?
» Must consider IMF’s
Formation of Solutions
•
Solute and solvent particles must
be attracted to one another:
“Like Dissolves Like”
•
For a solution for form…..
Solute-solvent IMF’s > Solvent-Solvent & solute-solute IMFS
The mode of
action of the
antibiotic,
Gramicidin A
Destroys the Na+/K+ ion
concentration gradients
in the cell
Figure B13.2
13-8
The arrangement of atoms in two types of alloys
Solid-solid solutions: alloys (substitutional or interstitial)
Figure 13.4
13-9
Solutions involving Liquids
•
•
Molecules of each liquid must be pushed
apart for a solution to form
Why doesn’t water form a solution with
n-Hexane, C6H14?
»
Water molecules too strongly attracted to each
other to be pushed aside to make room for
hexane molecules
Application Questions: Solutions involving Liquids
1.
Why does a solution form between water and
ethanol?
• H-Bonding between water and ethanol molecules is
responsible for solution formation
– Allows water molecules to be pushed aside to make
room for ethanol molecules
2.
Why do all nonpolar liquids mix to form
solutions?
» e.g. Oil and n-Hexane
SAMPLE PROBLEM 13.1
Predicting Relative Solubilities of Substances
Predict which solvent will dissolve more of the given solute:
(a) Ethylene glycol (HOCH2CH2OH) in hexane
(CH3CH2CH2CH2CH2CH3) or in water.
(b) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol
(CH3CH2OH)
(c) Sodium chloride in methanol (CH3OH) or in propanol
(CH3CH2CH2OH)
PLAN:
Consider the intermolecular forces which can exist between
solute molecules and consider whether the solvent can
provide such interactions and thereby substitute.
Solids Dissolved in Liquids
1.
2.
Solid particles must separate for a solution to form
(T2)
» Solute must be attracted to solvent
– Ion-Dipole attractions
– Solvation vs Hydration
What happens if the attractive forces within solute
and solvent differ greatly?
» e.g. Hexane and NaCl
Figure 13.1
The major
types of
intermolecular
forces in
solutions
Figure 13.2
Hydration
shells
around an
aqueous
ion
What determines if the DHsoln for a solid with a
liquid is exo- or endothermic? (T3-5)
1.
DHsoln = Lattice Energy + Solvation Energy
2.
Lattice Energy
»
3.
E needed to separate particles: Always
Endothermic
Solvation Energy
»
»
E released upon solvation
Always Exothermic
Figure 13.5
Solution cycles and the enthalpy components of the heat of solution
A. Exothermic Solution Process
B. Endothermic Solution Process
Heats of solution and solution cycles
1. Solute particles separate from each other - endothermic
solute (aggregated) + heat
solute (separated)
DHsolute > 0
2. Solvent particles separate from each other - endothermic
solvent (aggregated) + heat
solvent (separated)
DHsolvent > 0
3. Solute and solvent particles mix - exothermic
solute (separated) + solvent (separated)
solution + heat
DHsoln = DHsolute + DHsolvent + DHsolvation
DHSolv < 0
Figure 13.6
Heats of Solution: Dissolving ionic compounds in water
NaCl
NH4NO3
DHsoln = +
DHsoln = +
NaOH
DHsoln = -
Figure 13.7 Enthalpy diagrams for dissolving NaCl and octane in hexane
Table 13.3
Trends in Heats of Hydration for Various Ions
Heats of Solution: Application Questions
1.
2.
3.
4.
Why is the DHsoln always exothermic (negative) for
solutions between gases and liquids?
Why is the DHsoln = 0 for solutions between gases?
Why is the dissolving of Calcium Chloride, CaCl2,
in water exothermic?
» CaCl2 (along with NaCl) are used to salt roads
Why is the dissolving of ammonium nitrate,
NH4NO3, in water endothermic?
» NH4NO3 is used in chemical cold packs
The Effect of Temperature on
Solubility
Objectives:
» Describe the effect of temperature on
solubility of gases, liquids, and solids in
liquids
Solubility
1.
Equation describing a saturated solution at
equilibrium
Solute + Solvent
Saturated Solution
2.
Most Common Units:
» Mass solute/100 g solvent at a given
temperature
Figure 13.8
Equilibrium in a saturated solution
Solute (undissolved)
Solute (dissolved)
Figure 13.10
The relation
between
solubility and
temperature for
several ionic
compounds
Effect of Temp. on the
Solubility of a Gas in a Liquid
1.
2.
3.
Solubility of a gas always decreases as
Temp. increases. Why??
Le Chatelier’s Principle is used to predict
how an increase in temp. affects the
solubility of a gas in a liquid.
Recall DHsoln is exothermic for all gases in a
liquid:
Gas + Liquid
Gas dissolved + E
Solubility of a Gas in a Liquid:
Applications
1.
2.
Thermal Pollution  Decreases O2 Solubility
» Streams and Rivers
– Salmon/Trout habitat restoration
» Deep lakes
– Warm water at surface, cold water deep
Why are the richest fisheries in the coldest
waters of the world?
Solubility of a Gas in a Liquid:
Application Questions
1.
2.
3.
Why do bubbles form on the side of a glass of
water? What do these bubbles consist of?
Why do sodas get flat as they sit?
Why are some ice cubes clear, others cloudy?
» How are clear ice cubes made?
Henry's Law:
S
gas + solvent
solution
gas

P
gas
•
•
•
•
gas volume decreases
gas pressure increases
more collisions with liquid surface
gas solubility increases
Henry’s Law
The solubility of a gas (Sgas) is directly
proportional to the partial pressure of the
gas (Pgas) above the solution.
S
gas

P
or
gas
S
P
1
1

S
S
P
gas

k P
gas
2
2
kH = Henry’s law constant
for a gas; units of mol/L.atm
13-32
H
Implications for scuba diving!
Application of Henry’s Law
1.
2.
Why does a soda start to bubble immediately after
opening the bottle?
The solubility of methane , the chief component in
o
natural gas, in water at 20.0 C and 1.0 atm pressure
is 0.025 g/L. What will the solubility be at 1.5 atm
o
pressure and 20.0 C ?
» Answer: 0.038 g/L
Section 13.4
Concentrations of Solutions
1.
2.
Be able to convert from one concentration unit
to another:
• Molarity: Review Section 3.5
• Molality
• Mass Fraction and Mass Percent
• Mole fraction and Mole Percent
Practice!! Practice!! Practice!!
Concentrations of Solultions:
Molarity, M
Molarity = moles of solute divided by Liters of Solvent
M = mol solute / L of solution
• Used in stoichiometric calculations involving
solutions since V x M = moles
• Since Volume varies with temperature, Molarity
varies w/ temperature
Concentrations of Solutions:
Molality, m
1.
Molality =Moles of solute divided by kg of Solvent
m = mol solute / kg solvent
2.
Does not change with Temp.
– Used in BP elevation and FP depression
calculations
Molality Practice #1
Water freezes at a lower temperature when it
contains solutes.
a.
Calculate the number of grams of methanol,
CH3OH, needed to prepare a 0.250 m solution,
using 2000. grams of water. Methanol: 32.00 g/mol.
Ans. = 16.0 g methanol
b.
Calculate the Freezing point of 0.250 m Methanol.
Ans. = - 0.465 oC
FP Lowering
DTf = Kf m
DTf = amount FP is lowered
Kf = Freezing point depession constant
Kf is solvent Dependent
o
1.86 C/m for water
o
5.07 C/m for benzene
o
20.0 C/m for cyclohexane
Molality Practice #2
a.
If you prepare a solution by dissolving 4.00 g
of NaOH in 250. g of water, what is the
molality of the solution? NaOH: 40.00 g/mol
Ans. = 0.400 m
b.
At what temperature would the solution
boil?
o
Ans. = 100.2 C
BP Elevation
DTb = Kb m
• Kb = Boiling point elevation constant
• Kb is solvent Dependent
o
0.51 C/m for water
o
2.53 C/m for benzene
o
2.69 C/m for cyclohexane
Parts by Mass (or Percent by Mass)
1.
Mass of component divided by total mass of
solution
Mass % = (msolute / m solution) x100
2.
Also known as weight percent: w/w or % w/w
Mass Percent Practice #1
a.
b.
c.
How many grams of NaOH are needed to prepare 250.0g of
1.00% NaOH in water? Ans. = 2.50 g NaOH
How many grams of water are needed? Ans. = 247.5 g
How many mL of water at 20.0 oC are needed? Ans.= 250.455 mL
•
d.
dwater at 20.0 oC = 0.9882 g/mL
What is the molality of the solution?
Ans.= 0.2525 m NaOH
d.
What is the approximate FP of the solution?
Ans. = -0.939 oC
Mass Percent Practice #2
a.
Concentrated hydrochloric acid can be purchased
from chemical supply houses as a solution that is
37% HCl by mass. HCl = 36.46 g/mol
What mass of conc. HCl is needed to make 1.0 liter
of 0.1 M HCl?
Ans. = 9.854 or 10 g conc HCl
b.
How would you make the 0.1 M HCl solution?
Variations of % by Mass:
ppm and ppb
1.
2.
3.
Use ppm and ppb when concentrations of solute
are very low
Parts per million
»
ppm = mass fraction x 106
Parts per billion
»
ppb = mass fraction x 109
Concentration Unit Conversion Problems
1.
Strategies.........
a. Determine the Units of Concentration involved
– What are the units you are starting with?
– What are the units you are converting to?
b. Figure our what conversion factors are needed to
go get you to the desired units of concentration
Concentration Unit Conversion
Practice #1
a.
Calculate the molality 2.00 % NaCl.
NaCl = 58.4425 g/mol
Ans. = 0.349 m NaCl
b.
How would you prepare
• 1.00 liter of 2.00 % NaCl (w/v)?
• 500. mL of 2.00 % NaCl (w/v)?
• 250. mL of 2.00 % NaCl (w/v)?
Concentration Unit Conversion
Practice #2
l
Conc. hydrobromic acid can be purchased as
40.0% HBr. The density of the solution is 1.38
g/mL.
What is the molar concentration of 40.0%
HBr? HBr = 80.912 g/mol
Ans. = 6.82 M HBr
Colligative Properties
Properties of a solution that depend on
the number of solute particles, not on
their identity:
»Vapor Pressure Lowering
»Freezing Point Lowering
»Boiling Point Elevation
»Osmotic Pressure (will not cover)
Vapor Pressures of Solutions
Which is higher, the vapor pressure
of salt water or that of pure water?
Vapor Pressures of Solutions
1.
A solution has a lower vapor pressure than
that of the pure solvent (if the solute is
nonvolatile). Why?
2.
Solute particles impede evaporation, but
do not affect condensation
Raoult’s Law:
o
Psoln = (Xsolvent)(P solvent)
Mole Fraction
Mole fraction
• moles of component divided by total moles of
all components present in the mixture
Xa = na / (na + nb + nc +....)
•
Used in Raoult’s Law Calculations
Calculating the
Vapor Pressure of a Solution
Raoult’s Law
Psoln =
o
(Xsolvent)(P solvent)
Psoln = Vapor Pressure of Solution
Xsolvent = mole fraction of solvent
o
P solvent = Vapor pressure of the pure solvent
Raoult’s Law: Practice makes perfect?
Dibutyl phthalate, C16H22O4 (mw = 278 g/mol), is
an oil sometimes worked into plastic articles to
give them softness. It has a negligible vapor
pressure (P = 1 torr @ 148 oC).
What is the vapor pressure at 20.0 oC of a solution
of 20.0 g dibutyl phthalate in 50.0 g of octane, C8H18
(mw = 114 g/mol)?
1.
»
The vapor pressure of pure octane at 20.0 oC is 10.5 torr.
Ans. = 9.02 torr
BP Elevation and FP Depression
Objectives
» Explain the effect of a solute on the melting/freezing
point and boiling point of a solution
» Use F.P. depression and B.P elevation data to calculate
the molar mass of a compound.
BP Elevation and FP Depression
1.
Nonvolatile Solutes Lower the Vapor
Pressure of a Solvent (T11)
» Causes….
BP Elevation, DTb
FP Depression, DTf
» Colligative Properties: Depends on conc. of
solute, not identity of solute
BP Elevation
1.
DTb = Kb m
» Kb = Boiling point elevation
constant
» Kb is solvent Dependent
o
0.51 C/m for water
o
2.53 C/m for benzene
o
2.69 C/m for cyclohexane
FP Depression
DTf = Kf m
» Kf = Freezing point depession constant
» Kf is solvent Dependent
o
1.86 C/m for water
o
5.07 C/m for benzene
o
20.0 C/m for cyclohexane
Practice makes perfect..........
1.
A solution made by dissolving 3.46g of an unknown
o
compound in 85.0 g of benzene (Kf = 5.07 C/m, FP
o
o
= 5.45 C) froze at 4.13 C. What is the molar mass
of the compound? Answer: 157 g/mol
2.
At what temperature will a 10.0 % aqueous sucrose,
C12H22O11, solution boil?
Answer: 100.60 oC
Colligative Properties of Solutions of Electrolytes
Ionic Compounds (electrolytes) dissociate into ions
when dissolved in water.
•
NaCl(s)  Na+ (aq) + Cl - (aq)
1 mol
•

2 mol of Ions in solution
CaCl2(s)  Ca2+ (aq) + 2 Cl - (aq)
1 mol

3 mol of Ions in solution
• (NH4) 2SO4 (s)  2 NH4 +(aq) + SO42- (aq)
1 mol

3 mol of Ions in solution
Colligative Properties of Solutions of
Molecular Substances
1.
Nonelectrolytes: Molecules separate when
forming solutions
C12H22O11 (s)  C12H22O11 (aq)
2.
Weak Electrolytes: Incomplete ionization of
molecules in solution (e.g. acetic acid)
HC2H3O2 (aq)
H +(aq) + C2H3O2 1- (aq)
Predicting Freezing Points
Estimate the FP of the following aqueous solutions
o
DTf = Kf m Kf for water = 1.86 C/m)
1. 1.00 m sucrose, C12H22O11
2. 2.00 m sucrose, C12H22O11
3. 1.00 m NaCl
4. 1.00 m MgSO4
5. 1.00 m HCl
6. 1.00 m Acetic Acid, HC2H3O2
Why are the expected FP’s higher than the
observed values for some solutions?
Expected
1.
2.
3.
4.
5.
6.
1.00 m sucrose:
2.00 m sucrose:
1.00 m NaCl:
1.00 m MgSO4:
1.00 m HCl:
1.00 m HC2H3O2:
o
Observed
o
-1.86 C vs. -1.86 C
o
o
-3.72 C vs. -3.72 C
o
o
-3.72 C vs. -3.53 C
o
o
-3.72 C vs. -2.42 C
o
o
-3.72 C vs. -3.53 C
o
o
-1.86 C vs. -1.90 C
Effects of Interionic Attractions
1.
2.
Dissociation into ions is not 100%
Ion pairs exist in solution....Thus....
» Number of moles of ions in a 1.0 m
NaCl solution is not double of the
molality
» Causes DTb and DTf to be smaller
than expected
van’t Hoff factor, i
Gives an indication of the % Dissociation of ions
in solution
i
Dt
Dt
f (measured)
f (calculated as if a nonelectrolyte)
Use the FP’s to calculate the van’t Hoff factor
for each compound
1.
2.
3.
o
1.00 m NaCl:
FP = -3.53 C
o
1.00 m MgSO4:
FP = -2.42 C
o
1.00 m HC2H3O2: FP = -1.90 C
i
Dt
Dt
f (measured)
f (calculated as if a nonelectrolyte)
van’t Hoff factors for ideal solutions
If 100 % dissociation, then for
a. NaCl : i = 2
b. KCl :
i = 2
c. MgSO4 : i = 2
d. K2SO4 : i = 3
e. Na3PO4: i = 4
Van’t Hoff factor:
Expected vs. Observed
values
The End
Good luck on your
final exams!!!