Transcript hc2-chapter 12 powerpoint

Ch 12. Properties of Solutions;
Mixtures of Substances at the
Molecular Level
Brady & Senese, 5th Ed.
1
Index
12.1. Substances mix spontaneously when there is no energy
barrier to mixing
12.2. Heats of solution come from unbalanced intermolecular
attractions
12.3. A substance's solubility changes with temperature
12.4. Gases become more soluble at higher pressures
12.5. Molarity changes with temperature; molality, weight
percentages, and mole fractions do not
12.6. Solutes lower the vapor pressure of a solvent
12.7. Solutions have lower freezing points and higher boiling
points than pure solvents
12.8. Osmosis is flow of solvent through a semipermeable
membrane due to unequal concentrations
12.9. Ionic solutes affect colligative properties differently
than nonionic solutes
2
Mixing Processes
• Mixing occurs due to interaction between
molecules “like dissolves like”
• As partition is removed, molecules are able to
move freely and interact
• Mixed state is statistically more probable
12.1. Substances mix spontaneously when there is no energy barrier to mixing
3
The Process Of Dissolution
• Polar solutes dissolve in polar solvents
• Non-polar solutes dissolve in non-polar solvents
• Dipoles of solvent may induce dipoles in solute,
effecting dissolution
12.1. Substances mix spontaneously when there is no energy barrier to mixing
4
Miscibility of Liquids
• Liquids that can dissolve
in one another are
miscible, while insoluble
liquids are immiscible
• Ethanol and water are
miscible, while benzene
and water are not
12.1. Substances mix spontaneously when there is no energy barrier to mixing
5
Learning Check
Which of the following are miscible in water?
water
acetic acid
carbon
disulfide
ammonia
12.1. Substances mix spontaneously when there is no energy barrier to mixing
6
Your Turn!
Which of the following are likely to be miscible
with water?
A. CH3CH2CH2CH3
B. C6H6
C. CH3CO2H
D. All are expected to be miscible
12.1. Substances mix spontaneously when there is no energy barrier to mixing
7
Dissolution Of An Ionic Compound In
Water
• Positive end of the dipole
of the water surrounds the
anions of the ionic solid,
extracting them from the
lattice
• Negative end of the dipole
orients toward the cations,
surrounding and extracting
them from the lattice
12.1. Substances mix spontaneously when there is no energy barrier to mixing
8
Dissolution Of A Polar Compound In Water
Dipole of the water interacts with the oppositely
charged dipoles of the solid, extracting them from
the crystal
12.1. Substances mix spontaneously when there is no energy barrier to mixing
9
Enthalpy (Heat) Of Solution
• Heat of solution (Ηsoln ) is the energy
exchanged when a solute dissolves in a solvent at
constant pressure
• Enthalpy is a state function: the pathway can be
written in any way and the result will be the same
• When Ηsoln=0, solution is called an ideal
solution
12.2. Enthalpy of solution comes from unbalanced intermolecular attractions
10
Dissolution Of An Ionic Solid
• Visualized in steps:
• step1: ionic solid
breaks apart into vapor
phase lattice energy
(U)
• step 2: vapor phase
interacts with solvent
solvation energy
(ΔHsolv); if solvent is
water, (Ηhydration)
Ηsoln (ion in water)= U + Ηsolvation
12.2. Heats of solution come from unbalanced intermolecular attractions
11
Dissolution: Liquid In Liquid
• Step1: solute
expands
• Step2: solvent
expands
• Step 3 solute &
solvent mix
• If the
Ηsoln=0, we
have an ideal
solution
Ηsoln = Η1 + Η2 + Η3
12.2. Heats of solution come from unbalanced intermolecular attractions
12
Dissolution: Liquid in Liquid (Ideal)
12.2. Heats of solution come from unbalanced intermolecular attractions
13
Dissolution: Gas In Liquid
• step1: expansion of solvent
• step2: mixing
• Ηsoln = Η1 + Η2
12.2. Heats of solution come from unbalanced intermolecular attractions
14
Your Turn!
What factor does not affect the value of ΔHsoln ?
A. The polarities of solute and solvent
B. The size of the solute
C. The charge on the solute
D. The temperature of the solution
E. All affect the value
12.2. Heats of solution come from unbalanced intermolecular attractions
15
Saturated Solutions
• Solute is at equilibrium
with the dissolved solute
• Addition of more
dissolved solute results in
supersaturation and
precipitation of excess
solid
• The presence of less solute
than the solubility results
in an unsaturated
solution
12.3. A substance's solubility changes with temperature
16
Solubility Varies With Temperature
• Solubility may increase
or decrease with
increasing temperature
• The extent to which
temperature has an
effect is specific to the
solute and solvent
• Most gases are less
soluble in water at high
temperature, while
most solids are more
soluble
12.3. A substance's solubility changes with temperature
17
Case Study: Dead Zones
During the industrial revolution, factories were built on
rivers so that the river water could be used as a coolant
for the machinery. The hot water was dumped back
into the river and cool water recirculated. After some
time, the rivers began to darken and many fish died.
The water was not found to be contaminated by the
machinery. What was the cause of the mysterious fish
kills?
increased temperature lowered amounts of
dissolved oxygen
12.3. A substance's solubility changes with temperature
18
Effects Of Temperature On Solubility
• Solubility varies with temperature according to the
enthalpy of solvation
• The efficiency of a solvation process (K) depends on the
enthalpy (ΔH) in Joules, the ideal gas constant (R), and the
temperature (T) in Kelvin
• If the dissolution process is endothermic (ΔΗ is +),
increasing temperature results in greater efficiency
K1
 H 1
1
ln( ) 
(  )
K2
R T1 T 2
12.3. A substance's solubility changes with temperature
19
Your Turn!
The solubility of a substances increases with
increased temperature if:
A. ΔHsolution >0
B. ΔHsolution <0
C. ΔHsolution =0
12.3. A substance's solubility changes with temperature
20
Pressure Effects On Solubility Of Gases
• Cg=kHPg
 C = concentration of dissolved gas (M)
 kH = Henry’s Constant
 P = pressure applied to system (mm Hg)
kH
(M/mm Hg)
N2
8.42×10 -7
O2
1.66×10-4
CO2
4.48×10-5
• Gases are all more soluble at higher pressures (the
cause of “the bends”)
12.4. Gases become more soluble at higher pressures
21
Learning Check
What is the concentration of dissolved nitrogen in a
solution that is saturated in N2 at 2.0 atm
kH= 8.42×10 -7 (M / mm Hg)
•Cg=kHPg
•Cg= 8.42×10 -7 (M / mm Hg) × 2.0 atm × 760 mmHg/atm
•Cg=1.3 ×10-3 M
12.4. Gases become more soluble at higher pressures
22
Your Turn!
A.
B.
C.
D.
When you open a bottle of
seltzer, it fizzes. How should you
store it to increase the time before
it goes flat?
Heat it and pressurize it
Cool it and pressurize it
Heat it and reduce the pressure
Cool it and reduce the pressure
12.4. Gases become more soluble at higher pressures
23
Units Of Concentration
• Molarity (M) = moles solute / L solution
 changes with Temperature
• Molality (m) = moles solute/kg solvent
• mole fraction (X)
 X = moles component/ total moles
• Percent by mass (%)
 (mass solute / mass solution)*100
12.5. Molarity changes with temperature; molality, weight percentages, and mole
fractions do not
24
Units Of Very Low Concentrations
• Parts per million (ppm)
 μg solute/mL soln
• Parts per billion (ppb)
 ng solute/ mL soln
• In extremely dilute solutions mostly solvent is
present
• When the solvent is water (d≈1g/mL) thus for ppm
≈ μg solute/g soln
• 1/106 magnitude difference leads to the name 1 part
per 1 billion
12.5. Molarity changes with temperature; molality, weight percentages, and mole
fractions do not
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Organize Your Thoughts!
• All concentration units are a ratio of information
• Develop a sense of the data that you have available
Solute Solvent
Solution
Solution
Volume
Mass
Mole
Reference MM
g/mol
MM
g/mol
d (g/mL)
12.5. Molarity changes with temperature; molality, weight percentages, and mole
fractions do not
26
Learning Check: What Does Molarity Tell Us?
M=moles solute/L solution. What are the m, X, %
and ppm concentration of a 1.0M solution of
KCl with a density of 0.99 g/mL
m = 1.1 X = 0.019 % =7.5 ppm=7.5(104)
Solute Solvent Solution Solution
Volume
KCl
(H2O)
Mass (g)
Mole
Reference
74.55
1.0
74.5510
g/mol
915.44
50.815
18.0153
g/mol
990
51.815
1L
0.99 (g/mL)
12.5. Molarity changes with temperature; molality, weight percentages, and mole
fractions do not
27
Learning Check: What Does Molality Tell Us?
m=moles solute/kg solvent.What are the M, X, %
and ppm concentration of 1.0 m KCl with a
density of 0.98 g/mL
M = 0.91 X = 0.018 % = 6.9 ppm =6.9×104
Solute Solvent Solution Solution
Volume
KCl
(H2O)
Mass (g)
Mole
Reference
74.55
1.0
74.5510
g/mol
1000
55.51
18.0153
g/mol
1074.55
56.51
1096 mL
=1.096 L
0.98 (g/mL)
12.5. Molarity changes with temperature; molality, weight percentages, and mole
fractions do not
28
Learning Check: What Does Mole Fraction Mean?
Xsolute = moles solute/moles total. What are the M, m, % and
ppm concentration of a solution that has XKCl = 0.060
with a density of 0.87 g/mL
M =2.4 m = 3.5 % = 21 ppm =1.8×105
Solute Solvent Solution Solution
Volume
KCl
(H2O)
Mass (g)
Mole
Reference
4.473
16.93
21.403
0.060
0.94
1
74.5510
g/mol
18.0153
g/mol
0.87 (g/mL)
24.601 mL
=.024601 L
12.5. Molarity changes with temperature; molality, weight percentages, and mole
fractions do not
29
Learning Check: What Does % Mass Tell Us
%=(mass solute/mass solution) x 100. What are the M, m, X and
ppm concentration of a 1.05 % KCl solution with a density of
1.15 g/mL
M =0.162 m = 0.142 X = 2.26×10-4 ppm =1.21×104
Solute Solvent Solution Solution
Volume
KCl
(H2O)
Mass (g)
1.05
Mole
.0140843 54.9255
Reference
74.5510
g/mol
98.95
18.0153
g/mol
100
55.0664
86.957 mL
1.15 (g/mL)
=.086957 L
12.5. Molarity changes with temperature; molality, weight percentages, and mole
fractions do not
30
Your turn!
Which of the following corresponds to a 3.5M
solution of NaCl with a density of 0.997 g/mL?
A.
m
0.0035
XNaCl
0.074
%
21
B.
3.5
0.080
0.21
C.
0.0035
0.074
0.21
D.
3.5
0.074
21
MM H2O: 18.0153; NaCl: 58.443
12.5. Molarity changes with temperature; molality, weight percentages, and mole
fractions do not
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Raoult’s Law
• Vapor pressure of a liquid varies as a function of
purity
• X= mole fraction of solvent
 P0= vapor pressure of pure solvent
• Psolution=XsolventP0solvent
• Psolution=XAP0A+XBPB0
 Where A and B are both volatile components.
12.6. Solutes lower the vapor pressure of a solvent
32
Learning Check
The vapor pressure of 2-methylheptane is 233.95 torr at 55°C.
3-ethylpentane has a vapor pressure of 207.68 at the same
temperature. What would be the pressure of the mixture of
78.0g 2-methylheptane and 15 g 3-ethylpentane?
•Psolution=XAP0A+XBP0B
2-methylheptane
Chemical Formula: C8H18
Molecular Weight: 114.23
•mole 2-methylheptane : 78.0g/114.23
g/mol = 0.68283 mol
•mole 3-ethylpentane: 15g/100.2 g/mol
= 0.1497 mol
•X2-methylheptane=0.8202 ; X3-ethylpentane =10.8202 = 0.1798
3-ethylpentane
Chemical Formula: C7H16
Molecular Weight: 100.2
P  0.8202  233.95torr  0.1798  207.68torr
P = 230 torr
12.6. Solutes lower the vapor pressure of a solvent
33
Learning Check
The vapor pressure of 2-methyl hexane is 37.986 torr at
15°C. What would be the pressure of the mixture of
78.0g 2-methylhexane and 15 g naphthalene which is
nearly non-volatile at this temperature?
•Psolution=XsolventP0solvent
naphthalene
Chemical
Formula:
C10H8
Molecular
Weight:
128.17
•mol 2-methylhexane: 78.0g/100.2 g/mol
= 0.778443 mol
•mol naphthalene: 15 g/128.17 g/mol =
0.11703
•X2-methylhexane = 0.869309
•Psolution = 0.869309 ×37.986 torr
•P=33.02 torr
12.6. Solutes lower the vapor pressure of a solvent
2-methylhexane
Chemical Formula: C7H16
Molecular Weight: 100.2
34
Your Turn!
n-hexane and n-heptane are miscible in a large degree and
both volatile. If the vapor pressure of pure hexane is
151.28 mm Hg and heptane is 45.67 at 25º, which
equation can be used to determine the mole fraction of
hexane in the mixture if the mixture’s vapor pressure is
145.5 mm Hg?
A. X(151.28 mmHg)=145.5 mmHg
B. X(151.28 mmHg) + (X)(45.67 mm Hg) = 145.5 mmHg
C. X(151.28 mmHg)+(1-X)(45.67 mm Hg)=145.5 mm Hg
D. None of these
12.6. Solutes lower the vapor pressure of a solvent
35
Solute Effects On Phase Changes:
• Regardless of the identity of the dissolved particles, the
presence of an impurity will result in a change in the
boiling point and freezing point.
• The effect is solely dependent on the nature of the
solvent, a factor labeled K, and the concentration of
particles present (m)
• ΔT=mK
 boiling point elevation ΔT=Tmix-Tpure
 freezing Point Depression ΔT=Tpure-Tmix
12.7. Solutions have lower freezing points and higher boiling points than pure solvents
36
Effects Of Impurities On Phase Changes
12.7. Solutions have lower freezing points and higher boiling points than pure solvents
37
Some BP/FP Constants
Solvent
Normal Kbp
BP (°C) (°C/m)
Normal
FP(°C)
Kfp
(°C/m)
Water
100.00
+.51
0.0
1.86
Acetic Acid
1118.3
+3.07
16.6
3.57
Benzene
80.2
+2.53
5.45
5.07
Camphor
207
+5.611
178.4
37.7
Chloroform
61.20
+3.63
-
-
Cyclohexane
80.7
2.69
6.5
20.0
12.7. Solutions have lower freezing points and higher boiling points than pure solvents
38
Learning Check
According to the Sierra™ Antifreeze literature, the
freezing point of a 40/60 solution of sierra antifreeze
and water is -4 °F. What is the molality of the solution?
-4°F = 1.8 (°C) + 32
-20. °C
ΔTsolution  m  Kfp

1.86 C
(0  ( 20.) C  ? m 
m
11=m

12.7. Solutions have lower freezing points and higher boiling points than pure solvents
39
Learning Check:
In the previous sample of a Sierra™ antifreeze
mixture, 100 mL is known to contain 42 g of the
antifreeze and 60. g of water, what is the molar
mass of the compound found in this antifreeze if it
has a freezing point of -4°F?
from before:
-4°F = 1.8 (°C) + 32 =-20. °C
mol solute
 10.75 m
0.060kg solvent
ΔTsolution  m  Kfp
0.6452 mol solute
1.86 C

(0  ( 20.) C  ? m 
m
650 g/mol solute

10.75=m
12.7. Solutions have lower freezing points and higher boiling points than pure solvents
40
Learning Check:
In the previous sample of a Sierra™ antifreeze
mixture, the freezing point is -4°F? What will be
its boiling point?
from before:
ΔTFreezing Point Kfp

m
-4°F = 1.8 (°C) + 32 =-20. °C
ΔTBoiling Point Kbp
ΔTfreezingsolution  m  Kfp
and
ΔT boilingsolution  m  Kbp
If it is thesame concentration,
ΔTfreezingsolution
ΔT boilingsolution
m
Kfp
Kbp
(0  C - (-20  C)) 1.86  C/m


Tmix - 100 C
0.51  C/m
T=105 °C
12.7. Solutions have lower freezing points and higher boiling points than pure solvents
41
Your Turn!
Beer is known to be around a 5% ethanol (C2H5OH)
solution with a density of 1.05 g/mL. What is its
expected boiling point?( Kf=0.51°/m)
A. 100ºC
B. 101ºC
C. 102ºC
D. 103ºC
E. Not enough information given
MM: H2O=18.0153; C2H5OH=46.069
12.7. Solutions have lower freezing points and higher boiling points than pure solvents
42
Osmosis
• When two solutions are separated by a semi-permeable
membrane, solvent molecules flow from areas of low
concentration to areas of high concentration
• As this occurs, the height of liquid rises in the higher
concentration solution, building up “Osmotic pressure”
(π)
12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal
concentrations
43
Relative Concentration Terms In Osmosis
• Hypotonic solutions have lower ion
concentrations than the cells.
• Hypertonic solutions have higher ion
concentrations than the cells
• Isotonic solutions have the same ion
concentration as the cells
12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal
concentrations
44
Osmosis
• π=MRT
 the concentration, is in molarity, M
 T=Temperature, in Kelvin
 R=Ideal Gas Constant, 0.082057 L·atm/mol·K
• The basis for kidney function, rising sap, and
dialysis
12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal
concentrations
45
Learning Check: Osmosis
A solution of D5W, 5% dextrose (C6H1206) in water is
placed into the osmometer shown at right. It has a
density of 1.0 g/mL. The surroundings are filled
with distilled water. What is the expected osmotic
pressure at 25°C?
  MRT
5g C6H12O6 1.0g soln mol C6H12O6 1000mL



M
100g solution mL soln
180.16g
L
0.277 mol 0.082057 L  atm


 298K
L
mol  K
  7 atm
12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal
concentrations
46
Learning Check
For a typical blood plasma, the osmotic pressure at body
temperature (37°C) is 5409 mm Hg. If the dominant
solute is serum protein, what is the concentration of
serum protein?
  MRT
5409 mm Hg
1atm


760 mm Hg
? mol 0.082057 L  atm
7.117atm 

 310.15K
L
mol  K
0.280 mol
M
L
12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal
concentrations
47
Dialysis
• Pores on the semi-permeable membrane may be of
varied size
• In dialysis, the pores are fairly large, allowing transfer
of solvent, ions, and small proteins
• Larger cells, such as red blood cells are prevented
from passing through the pores
• The dialysis bath may be enriched in substances
lacking in the blood, and is hypotonic in waste
products in the blood
• Exchange of vital components may be made
12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal
concentrations
48
Your Turn!
Suppose that your tap water has 250 ppb of dissolved
H2S , and that its density is about 1.0 g/mL. What
is its osmotic pressure at 25°C?
A. 0.00058 atm
B. 0.064 atm
C. 0.059 atm
0.21 atm
D. None of these
MM: H2S =34.076
12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal
concentrations
49
Ionic Solutes Affect Colligative Properties
Differently Than Non-ionic Solutes
• substances that ionize make more particles in a solution
than their own concentration suggests
• i is a factor that demonstrates how many ions are formed
per formula unit or molecule
• the apparent molality of particles is then im.
ΔTmeasured
ΔTcalculatedassuming no ionization
Tmeasured
i
Tcalculated

imK
mK
12.9. Ionic solutes affect colligative properties differently than nonionic solutes
50
Learning Check
In preparing pasta, 2 L of water at 25°C are combined
with about 15 g salt (NaCl, MM= 58.44g/mol) and
the solution brought to a boil. What is the expected
boiling point of the water?
ΔT=imKbp
mass of water =volume ×density
=2000 mL ×1.0 g/mL
m=0.25667 mol / 2kg
=2000g water = 2 kg
=0.123
mol NaCl = 15g / 58.44 g/mol
mol NaCl = 0.25667
T  100  C 
2 ion 0.123m 0.51  C


mol
1
m
T=100.1 °C
12.9. Ionic solutes affect colligative properties differently than nonionic solutes
51
Case Study
Suppose you run out of salt. What mass of sugar
(C12H22O11, MM=342.30 g/mol) added to 2 L of
water would raise the temperature of water by 0.10
°C?
mass of water =volume ×density
=2000 mL ×1.0 g/mL
=2000g water = 2 kg
0.196 m=? mol / 2kg
0.39215mol
ΔT=imKbp
1 molec
0.51 C

0.10 C 
?m
mol
m
m=.196
0.39215 mol = ?g / 342.30 g/mol
mass sucrose =130 g
12.9. Ionic solutes affect colligative properties differently than nonionic solutes
52