DE Chemistry Dr Walker Solutions In a solution the excess material is the solvent the smaller amount is the solute Measuring Solution Composition Molarity is most common covered previously in Chapter 4 ID: 935310
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Slide1
Chapter 11Properties of Solutions
DE Chemistry
Dr. Walker
Slide2Solutions
In a solution, the excess material is the
solvent
, the smaller amount is the solute
Slide3Measuring Solution Composition
Molarity is most common, covered previously in Chapter 4
Slide4Measuring Solution Composition
Mass Percent
Slide5Mass Percent Example
What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol
(C
2H
5
OH) in
85.0
mL
of water
?
Slide6Mass Percent Example
What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol (C
2
H
5
OH) in 85.0 mL of water?
Density of water = 1.0 g/mL, therefore 85 mL of water = 85 g
Mass of solute/mass of solution x 100 = mass %
5.3 g ethanol/(5.3 g ethanol + 85.0 g water) =
5.9%
Slide7Measuring Solution Composition
Mole Fraction
Slide8Mole Fraction Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C
2
H5
OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Slide9Mole Fraction Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C
2
H
5
OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Calculate moles of ethanol:
5.30
g
C
2
H
5
OH
1 mole
C
2
H
5
OH
46.08
g
C
2
H5OH
=
0.115
moles
C
2
H
5
OH
Mole Fraction Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C
2
H
5
OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Calculate moles of water:
85.0
g H
2
O
1 mole H
2
O
=
4.72 moles
H
2
O
18.02 g H
2
O
Slide11Mole Fraction Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C
2
H
5
OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Mole Fraction of ethanol
Mole Fraction of Water
0.115
moles
C
2
H
5
OH + 4.72 moles
H
2
O
0.115
moles
C
2H5OH
= 0.024
0.115
moles
C
2
H
5
OH
+
4.72
moles H
2
O
4.72
moles H
2O
=
0.976
Slide12Measuring Solution Composition
Molality
Slide13Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C
2
H
5
OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Slide14Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C
2
H
5
OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Molality = moles solute/kg solvent
5.30
g
C
2
H
5
OH
1 mole
C
2
H
5
OH
46.08
g
C
2
H
5
OH
=
0.115
moles
C
2
H
5
OH
Measuring Solution Composition
Normality
Equivalents of acids and bases
Mass
that donates or accepts a mole of
protons (usually a subscript of H
+
or OH
-
)
Equivalents
of oxidizing and reducing agents
Mass
that provides or accepts a mole of
electrons (usually a subscript)
Slide16Normality
Find molarity first
Normality = Molarity x number of acidic protons
1.5 M
HCl
= 1.5 N
HCl
(1 acidic proton)
1.5 M H
2
SO
4
= 3.0 N H
2
SO
4
(2 acidic protons)
1.5 M H
3
PO
4
= 4.5 N H
3
PO4 (3 acidic protons)
Slide17Energy of Solution Formation
“Like Dissolves Like”
Polar molecules and ionic compounds tend to dissolve in polar solvents
Nonpolar molecules dissolve in nonpolar compounds
Water - polar
Butane - nonpolar
Slide18Heat of Solution
The
Heat of Solution
is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.
Substance
Heat of Solution
(kJ/mol)
NaOH
-44.51
NH
4
NO
3
+25.69
KNO
3
+34.89
HCl
-74.84
Slide19Steps In Solution Formation
Slide20Steps in Solution Formation
H
1
Step 1 -
Expanding the solute
Separating the solute into individual components – usually endothermic
Slide21Steps in Solution Formation
H
2
Step 2 -
Expanding the solvent
Overcoming intermolecular forces of the solvent molecules – usually endothermic
Slide22Steps in Solution Formation
H
3
Step 3 -
Interaction of solute and solvent to form the solution – usually exothermic
Slide23Enthalpy of Solution
May be positive or negative
Negative is favorable (exothermic)
Positive values are endothermic, does not dissolve spontaneously
Solution formation increases entropy (favorable)
Slide24Factors Affecting Solubility
Structure Effects
Polar (hydrophilic) dissolves in polar
Water soluble vitamins
Nonpolar (hydrophobic) in nonpolar
Fat soluble vitamins
Slide25Pressure Effects – Henry’s Law
The concentration of a dissolved gas in a solution is directly proportional to the pressure of the gas above the solution
Applies most accurately for dilute solutions of gases that do not dissociate or react with the solvent
Yes
CO
2
, N
2
, O
2
No
HCl
, HI
remember, these dissociate in solution!
Slide26Henry’s Law Example
Slide27Henry’s Law Example
Unopened bottle:
C =
k
P C = (3.1 x 10
-2
mol/L
.
Atm)(5.0 atm) =
0.16 mol/L
Opened bottle:
C =
k
P C = (3.1 x 10
-2
mol/L
.
Atm)(4.0 x 10
-4
atm) =
1.2 x 10
-5
mol/L
Slide28Solubility – Temperature Effects
Solids
Increases in temperature always cause dissolving to occur more rapidly
Increases in temperature usually increases solubility (the amount that can be dissolved)
Gases
Solubility of gases always decreases with increasing temperatures
Slide29Vapor Pressure – Nonvolatile solutes
Nonvolatile electrolytes lower the vapor pressure of a solute
Nonvolatile molecules do not enter the vapor phase
Fewer molecules are available to enter the vapor phase
Dissociation of ionic compounds has nearly two, three or more times the vapor pressure lowering of nonionic (nonelectrolyte) solutes.
Slide30Slide31Raoult’s Law
The presence of a nonvolatile solute lowers the vapor pressure of the solvent.
P
solution
= Observed Vapor pressure of
the solution
P
0
solvent
= Vapor pressure of the pure solvent
solvent
= Mole fraction of the solvent
This should make sense, since you’ve learned previously that dissolved
Solutes
raise
boiling points. Lower vapor pressure =
higher
boiling point
Slide32Raoult’s Law Example
1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25
o
C. The vapor pressure of water alone is 23.8 mm Hg at 25
o
C. What is the new vapor pressure of Kool-Aid?
Slide33Raoult’s Law Example
1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25
o
C. The vapor pressure of water alone is 23.8 mm Hg at 25
o
C. What is the new vapor pressure of Kool-Aid?
Usually, you must solve for the mole fraction of the solvent first:
2000 g H
2
O
1 mole H
2
O
18.02 g H
2
O
= 110.9 moles H
2
O
Mole Fraction of water (remember 1 L = 1000 g)
1.5 moles kool aid + 110.99 moles H
2
O
110.99 moles H
2
O
= 0.987
Slide34Raoult’s Law Example
1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25
o
C. The vapor pressure of water alone is 23.8 mm Hg at 25
o
C. What is the new vapor pressure of Kool-Aid?
Using the mole fraction from the previous page
P
Kool-Aid
=
c
H2O
P
pure H2O
= (.987)(23.8 mm Hg) = 23.5 mm Hg
Slide35Slide36Liquid-liquid solutions in which both components are
volatile (non-ideal solutions)
Modified Raoult's Law:
P
0
is the vapor pressure of the pure solvent
P
A
and
P
B
are the partial pressures
This is a variation on Dalton’s Law of Partial Pressures that accounts for
Raoult’s
Law
Slide37Modified Raoult’s Law Example
What is the vapor pressure
of the solution?
Slide38Modified Raoult’s Law Example
Slide39Modified Raoult’s Law Example
What is the vapor pressure
of the solution?
Slide40Ideal Solutions
Liquid-liquid solution that obeys Raoult’s Law
Like gases, none are ideal, but some are close
Negative Deviations
Lower than predicted vapor pressure
Solute and solvent are similar, strong forces of attraction
In a solution of acetone and
ethanol, there is a negative
deviation due to the hydrogen
bonding interactions shown
Slide41Ideal Solutions
Positive Deviations
Higher than predicted vapor pressure
Particles easily escape attractions in solution to enter the vapor phase
Ethanol and hexane are not attracted to each other due to differences
In polarity. As a result, its solution is a positive deviation (higher
D
H)
From Raoult’s Law.
Slide42Colligative Properties
Properties dependent on the number of solute particles but not on their identity
Boiling-Point elevation
Freezing-Point depression
Osmotic Pressure
Slide43Boiling Point Elevation
Each mole of solute particles raises the boiling point of 1 kilogram of water by 0.51 degrees Celsius.
K
b
= 0.51
C kilogram/mol
m
=
molality
of the solution
i
=
van’t
Hoff
factor
Slide44Freezing Point Depression
Each mole of solute particles lowers the freezing point of 1 kilogram of water by 1.86 degrees Celsius.
K
f
= 1.86
C kilogram/mol
m
=
molality
of the solution
i
=
van’t
Hoff
factor
Slide45Boiling Point Elevation
A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C
2
H6
O
2
, d=1.11 g/mL) and water is used, what will be the new boiling point?
Slide46Boiling Point Elevation
A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C
2
H6
O
2
, d=1.11 g/mL) and water is used, what will be the new boiling point?
Van’t Hoff factor = 1 (not an ionic salt)
K
b
= 0.512 kg/mol
Molality = moles/kg
3000 mL C
2
H
6
O
2
1.11 g C
2
H
6
O
2
1 mL C
2
H
6
O
2
1 mole C
2
H
6
O
2
62.08 g C
2
H
6
O
2
=
53.15
moles
Slide47Boiling Point Elevation
A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C
2
H6
O
2
, d=1.11 g/mL) and water is used, what will be the new boiling point?
Van’t Hoff factor = 1 (not an ionic salt)
K
b
= 0.512 kg/mol
Molality = moles/kg
3 L water = 3 kg water
m = moles solute/kg solvent
m =
53.16
moles C
2
H
6
O
2
/3 kg water =
17.72
m
DT = i . Kb. m= 1 x 0.512 x 17.72 = 9.15 C + 100 = 109.07 degrees C
Slide48Freezing Point Depression
Give the freezing point depression from adding 10 g ethylene glycol to 100 g water.
DT = (1) (1.86) (m)
10 g C2H6O2 / 620.8g g/mol = 0.16 moles
m = 0.16 moles/0.1 kg = 1.6 m
DT = (1)(1.86)(1.6) = 2.88 degrees
Slide49Determination of Molar Mass by Freezing Point Depression
Slide50Determination of Molar Mass by Freezing Point Depression
Rearrange equation
M
solute
=
D
T/K
f
M
solute
= 0.240 C / 5.12 C
.
kg/mol = 4.69 x 10
-2
mol/kg
Slide51Determination of Molar Mass by Freezing Point Depression
Rearrange equation
M
solute
=
D
T/K
f
M
solute
= 0.240 C / 5.12 C
.
kg/mol = 4.69 x 10
-2
mol/kg
0.015 kg benzene
Moles hormone
=
4.69 x 10
-2
mol/kg
Slide52Determination of Molar Mass by Freezing Point Depression
0.015 kg benzene
Moles hormone
=
4.69 x 10
-2
mol/kg
Molality
Moles hormone = 7.04 x 10
-4
moles
Determine molar mass
Slide53Van’t Hoff Factor, i
For ionic compounds, the expected value of
i
is an integer greater than 1
NaCl,
i
= 2
BaCl
2
,
i
= 3
Al
2
(SO
4
)
3
,
i
= 5
Total number of ions in solution
Slide54Dissociation Equations and the Determination of
i
NaCl(s)
AgNO
3
(s)
MgCl
2
(s)
Na
2
SO
4
(s)
AlCl
3
(s)
Na
+
(aq) + Cl
-
(aq)
Ag
+
(aq) + NO
3
-
(aq)
Mg
2+
(aq) + 2 Cl
-
(aq)
2 Na
+
(aq) + SO
4
2-
(aq)
Al
3+
(aq) + 3 Cl
-
(aq)
i
= 2
i
= 2
i
= 3
i
= 3
i
= 4
Slide55Freezing Point Depression and Boiling Point Elevation Constants,
C/
m
Solvent
K
f
K
b
Acetic acid
3.90
3.07
Benzene
5.12
2.53
Nitrobenzene
8.1
5.24
Phenol
7.27
3.56
Water
1.86
0.512
Slide56Osmotic Pressure
Semipermeable Membrane
Membrane which allows
solvent but not solute
molecules to pass through.
As time
passes…
One side of the membrane is mostly solvent
The other side
(which didn’t pass through) is more concentrated since the solute can’t go through the membrane
Osmosis
The flow of solvent into the solution through the semipermeable membrane
Slide57Osmotic Pressure
The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution
http://chemwiki.ucdavis.edu/Textbook_Maps/General_Chemistry_Textbook_Maps/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.07%3A_Osmotic_Pressure
Slide58Osmotic Pressure Calculations
= Osmotic pressure
M
=
Molarity
of the solution
R
= Gas Constant = 0.08206
Latm
/
molK
i
=
van’t
Hoff Factor
See page 509 in text for example
Slide59Example
Slide60Dialysis
Transfer of solvent molecules as well as small solute molecules and ions
Remember in osmosis, only solute molecules are transferred
This description fits the “dialysis” used to filter the blood when the kidneys do not work properly.
Slide61Kidney Dialysis
Dialyser contains ions and small molecules in blood, but none of the waste products.
Slide62Osmotic Pressure and Living Cells
Crenation
Cells placed in a hypertonic (higher osmotic pressure) solution lose water to the solution, and shrink
Hemolysis
Cells placed in a hypotonic (lower osmotic pressure) solution gain water from the solution and swell, possibly bursting
Slide63Reverse Osmosis
External pressure applied to a solution can cause water to leave the solution
Concentrates impurities (such as salt) in the remaining solution
Pure solvent (such as water) is recovered on the other side of the semipermeable membrane
Applicable to desalination plants which can make drinking water from ocean water.
Slide64Colloids
Tiny particles suspended in some medium
Particles range in size from 1 to 1000 nm
Noticeable by shining light through the mediumParticles are large enough that they scatter light
Slide65Examples of Colloids
Slide66Tyndall Effect
Scattering of light by particles
Light passes through a solution
Light is scattered in a colloid
http://www.dorthonion.com/drcmcm/CHEMISTRY/Lessons/Lectures/images/14TyndallEffect.jpg