Note to alanah – change henry’s law to be consistent with how the world
Uses it as evaporation
Hydrophobicity affects evaporation as molecule moves to get out of water
Environment.
Raoult’s Law
k H  M   PA
evaporation
PA   A PAo
Pao
Vapor pressure
 g 
Po
 MM  o  mole  o  L  o
P  
k Hcalculated  
P  
P 
Note check this
g
 s 


mole
[max imum M ]


 L 
Predict: high water interaction low evaporation
low vapor pressure (strong intermolecular interactions) low evaporation
Evaporation can be considered a two step process

aq 
A
1
s
Aliquid
g/L; kg/L; mg/L
Related to strength of dipole interactions with water
Inverse of Water solubility
Aliquid

aq 
A


V po
Agas
Related to intermolecular interactions of
The liquid to liquid to hold it as liquid,
Dispersion forces
V po
Agas
s
V po
Aaq 
 Agas
 g   mole 
 

 L  g 

Convert to concentration (M)
 MM V po
s
 k Henry
MAJOR LAWN PESTICIDES
HERBICIDES
Atrazine(1,2)
Balan(1)
Betasan(1)
2,4-D(1,2)
Dacthal(1,2)
Dicamba(1,2)
DSMA
Endothall(1)
Glyphosate(1,2)
MCPA(1,2)
MCPP
MSMA
Oxadiazon(1) Pronamide (1,2) Siduron
INSECTICIDES
Acephate(1,2) Baygon
Bendiocarb(1,2)
Carbaryl(1,2)
Chlorpyrifos(1,2) DDVP(1)
Diazinon(1)
Malathion(1,2)
Methoxychlor(1)
Oftanol(1)
Trichlorfon(2)
Triumph(1)
FUNGICIDES
Bayleton(1)
Benomyl.(1,2)
Chlorothalonil(1,2) Diphenamid(1,2)
Maneb(1)
PCNB(1,2)
Sulfur(1,2)
Ziram(1)
Balan
Dacthal
Oxadiazon
2,4-D
MCPP
Pronamide
Betasan
Dicamba
MCPA
Siduron
Atrazine
MSMA
Endothall
Glyphosate
Context Slide: Preparation for your papers
Reference Data for Risk Assessment
KOW 
Cg  kPg
g solute / Loc tan ol
g solute / Lwater
Your paper
for a compound of interest
s
g solute
g H2O
find, listing source of data
s
Kow
KH
octanol
H3C
Set a reference state to compare all chemicals
solubility in water (s)
solubility between water and octanol (Kow)
KH
OH
Why would this
Be important?
Hint, think about
Vit. C and Vit A
http://chem2.sis.nlm.nih.gov/chemidplus/
http://webbook.nist.gov/chemistry/
carbons
name
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
NIST
molar mass
g/mole
methane
16.0416
ethane
30.0674
propane
44.0932
butane
58.119
pentane
72.1448
hexane
86.1706
heptane
100.1964
octane
114.2222
nonane
128.248
decane
142.2738
undecane
156.2996
n-dodecane
170.3254
tridecane
184.3512
tetradecane
198.377
pentadecane 212.4028
hexadecane 226.4286
heptadecane 240.4544
octadecane
254.4802
nonadecane
268.506
eicosane
282.5318
heneicosane 296.5576
n-docosane
310.5834
n-tricosane
324.6092
n-tetracosane 338.635
n-pentacosane 352.6608
n-hexacosane 366.6866
n-heptacosane 380.7124
octacosane
394.7382
n-nonacosane 408.764
n-triacontane 422.7898
n-hentriacontane
436.8156
b.p.
deg C
(NIH site)
(NIST site)
Here is some data I gathered
A year ago for an env. project
water/pure substance
octanol/water gas/water
Solubility
log P
Henry's Law
mg/L (25C)
neg log solubility
atm-m3/mole
-162
22 -1.34242
-88.6
60.2 -1.7796
-42.1
62.4 -1.79518
-0.5
61.2 -1.78675
36
38 -1.57978
68.7
9.5 -0.97772
98.5
3.4 -0.53148
125.6
0.66 0.180456
150.8
220 -2.34242
174.1
0.052 1.283997
195.9
0.0044 2.356547
216.3
0.0037 2.431798
235.4
0.0047 2.327902
253.5
0.0022 2.657577
270.6
0.000076 4.119186
286.8
0.0009 3.045757
302
0.000294 3.531653
316.3
0.006 2.221849
329.9
0.0000297 4.527244
343
0.0019 2.721246
356.5 0.000000029 7.537602
368.6
7.9E-09 8.102373
380 0.000000295 6.530178
391.3
9.25E-08 7.033858
401.9 0.000000029 7.537602
412.2
0.0017 2.769551
442
2.83E-09 8.548214
431.6
8.84E-10 9.053548
440.8
2.76E-10 9.559091
449.7
8.58E-11 10.06651
458
2.67E-11 10.57349
1.09
1.81
2.38
2.89
3.39
3.9
4.66
5.18
4.76
5.01
6.5
6.1
6.73
7.2
7.71
8.25
8.69
9.18
9.67
10.16
10.65
11.15
11.64
12.13
12.62
13.11
13.6
14.09
14.58
15.07
15.57
gas over pure
gas/pure
substance
substance
mm
atm
0.658
466000
0.5
31500
0.707
7150
0.95
1820
1.25
514
1.8
151
2
46
3.21
14.1
3.4
4.45
5.15
1.43
1.93
0.412
8.18
0.135
2.889
0.0558
9.2
0.0116
12.6 0.00343
0.473 0.00143
38.5 0.000228
51.2 0.000341
67.9 0.000049
90.2 4.62E-06
120 8.73E-05
159 1.28E-06
211 1.75E-05
280 4.07E-06
372 1.51E-06
494 4.68E-07
655 2.81E-07
870
1.6E-09
1150
4.3E-10
1530 2.73E-11
2040
1.4E-11
613.1579
41.44737
9.407895
2.394737
0.676316
0.198684
0.060526
0.018553
0.005855
0.001882
0.000542
0.000178
7.34E-05
1.53E-05
4.51E-06
1.88E-06
3E-07
4.49E-07
6.45E-08
6.08E-09
1.15E-07
1.68E-09
2.3E-08
5.36E-09
1.99E-09
6.16E-10
3.7E-10
2.11E-12
5.66E-13
3.59E-14
1.84E-14
Solubility should be large with large number of polar or ionized groups
should decrease with increasing number of C groups
12
10
C21
-log (Solubility)
8
6
C8
C12
c15
4
greater solubility
2
C18
C20
C26
0
0
50
100
150
200
250
300
350
400
450
500
-2
C9
Increasing length of carbon chain
-4
molar mass (g/mole)
For comparison at 25oC
NaNO3
CaCl2
Pb(NO3)2
KNO3
NaCl, KCl
K2Cr2O7
Ce2(SO4)3
g/100g
92
92
62
38
32
15
5
g/1000g
9.2
9.2
6.2
3.8
3.2
1.5
0.5
-log s
-0.963788
-0.963788
-0.792392
-0.579784
-0.50515
-0.176091
0.30103
We can conclude from this data and the preceding slide that a –log s value of -0.5
Is pretty soluble (table salt)
In data collected from NIH and NIST kH is not in the same form as we had in Gen Chem
P
 L  atms 
kH  
 
 mole  M olarity
It represents the reaction in which a gas is released from the aqueous phase
Ag ,aq 
 Ag ,atm
In this context a large kH means that it will easily volatilize (escape from water)
Cg  kPg
In Gen Chem, the larger k,
The more it goes into the aqueous
Phase.
True for NIST and NIH values
Henry’s law can be calculated from the aqueous solubility (meaning how well it
interacts with polar molecules of water) and it’s vapor pressure (meaning how well it
interacts with itself, sufficient to stay in a liquid phase).
 MM 
kH  
V
 s  p
From this we would conclude the larger vapor pressure over
a pure liquid the larger Henry’s constant and the more likely
the compound is to escape as a gas from water. The greater
the molar aqueous solubility (molar mass over solubility
MM/s) the lower kH – The greater the number of polar groups
the larger the molar solubility and therefore the more it
interacts with water and the less it escapes
Log Kow >5 considered
A bioaccumulator
Log Octanol-Water Partitioning Coefficient (log Kow)
The octanol-water partition coefficient (Kow) is a measure of the equilibrium
concentration of a compound between octanol and water that indicates the
potential for partitioning into soil organic matter (i.e., a high Kow indicates a
compound which will preferentially partition into soil organic matter rather
than water). Kow is inversely related to the solubility of a compound in water.
Log Kow is used in models to estimate plant and soil invertebrate
bioaccumulation factors.
Chemicals with low Kow values (e.g., less than 10) may be considered relatively
hydrophilic; they tend to have high water solubilities, small soil/sediment
adsorption coefficients, and small bioconcentration factors for aquatic life.
Conversely, chemicals with high Kow values (e.g., greater than 104) are very
hydrophobic.
Conclude:
log KOW  1 ~ hydrophilic; high s
log KOW  104 or105 ~ hydrophobic, low s, bioaccumulator
Should there be a relationship between kH and KOW?
 MM 
kH  
V
 s  p
Alanah No
k OW
dispersion

polarity
We can say that the lower the dispersion forces the higher the Vp (and the lower
The boiling point). We know that as mass of the molecule increases the number of
electrons increase and so there should be some corresponding increase in the
dispersion forces so we might predict:
MM  dispersion
Which would allow us to predict, if everything else is equal (HA!)
 s
k OW  polarity  dispersion  MM    k H
 VP 
 s polarity 
This kind of reasoning suggests some


k OW  
k H Relationship between the two constants

VP


Which is shown for similar shaped compounds
On next slide
6
5
Kow
4
n-alkanes
n-chloromethane
3
2
1
0
0
1
2
3
Henry's Constant (L-atm/mol)
4
Chemistry 102-005
Example Paper turned in today
18 April 2007
Delta-9-Tetrahydrocannabinol (C21H20O2), or THC, is the pharmaceutical
compound found in the cannabis plant. The smoking of cannabis gives users a feeling of
euphoria, which is a result of the THC bonding to the brain’s cannabinoid receptors.
THC’s Kow of 6.97 indicates that it is generally insoluble in water. This is because THC
only has one OH- functional group, which generally dictates a molecule’s solubility in
water. Additionally, THC has a t1/2 of 57 hours because the compound is insoluble in
water, stored in the fatty tissues in the body, and is not easily secreted.
Δ9-Tetrahydrocannabinol (C21H20O2)
KOW
g solute / Loc tan ol

g solute / Lwater
From NIH: log Kow = 7.6
S = 2800 mg/L = 2.8ug/mL
CAS: 1972-08-3
LD50m mouse: 12.58 mg/kg
Example Paper turned in today
Ammonia (NH3) is a toxic gas that otherwise has many industrial and agricultural
uses. Ammonia serves a very important purpose in producing nitric acid by means of a
catalyst and exposure to high temperatures. After oxidation to nitric acid, NO, it's reacted
with water and can be used to provide nutrition to plants as a fertilizer as well as an
ingredient in explosives. Ammonia is extremely water soluble, indicated by its low Kow
of 1.69, because the central nitrogen atom has an empty electron pair and readily bonds
Wikepedia: water solubility: 89.9 g/100
mL 0oC
Cg  k H PA
with H+ ions in water to form NH4+.
NH3 + H2O ? OH- + NH4
Also, NH3 has a very small Henry’s Law constant (1.6 * 10-5), which indicates that not a
lot of heat is required to make ammonia to react.
Note Henry’s constant reported
differently by different groups – list with
units for interpretation
Doesn’t show up in NIH: NIST = between 55 and 61 mol/kg-bar
CAS registry: 7664-41-7
Google KOW ammonia = EPA superfund pdf document:
Ammonia (NH3)
ammonia
“A” students work
(without solutions manual)
~10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th & F 2-3:30 pm
Module #13
Solution Properties
Defining types of concentrations
For mixtures
SOLUTIONS
solvent
solute
Define some measurement scales and units
1.
Molarity
2.
mole fraction
3.
molality
M
G3: Science Is referential
molessolute
Lsolution
nA
 A
n A  nB .
molessolute
molality  m 
kg solvent
Method for Conversion from Molarity to molality
1. Assume 1 L volume
2. Calculate moles of solute
 molessolute 

 1L  molessolute
 Lsolution 
3. Calculate g of solution from d, V
 103 mL 
d solution 
 1Lsolution   g solution
 L 
4. Calculate g of solute from M
 g solute 
  g solute
M 1Lsolution 
 molsolute 
5. Subtract to get g of solvent
g solution  g solute  g solvent
6. Calculate g solute/g solvent
molessolute
molality  m 
kgsolvent
Example Density of an aqueous solution of ammonium sulfate is 1.06g/mL and the
molarity is 0.886. What is the molality?
NH SO


4 2
1. Assume 1 L volume
2. Calculate moles of solute
 molessolute 

 1L  molessolute  0.886
 Lsolution 
3. Calculate g of solution from d, V
4
Atoms
amu
total
2N
8 H 
S
4 O
total
14.01
81008
. 
32.07
416.00
28.02
8.064
32.07
64
133.098
g   1000mL 

.
 106

 L
  1060g solution

mL  
L  solution
4. Calculate g of solute from M
 133.098g solute 
mol 

  117.9248g solute
 0.886
 1Lsolution 

L 
 molsolute 
5. Subtract to get g of solvent
1060 g solution  117.9248 g solute  942.1g soluent
6. Calculate g solute/g solvent
molessolute 0.886molessolute
molality  m 

 0.941
kg solvent
0.9421kg solvent
Method for Conversion from molality to Molarity
1. Assume 1 kg solvent
2. Calculate moles of solute
 molessolute 

 1kg solvent  molessolute
 kg solvent 
3. Calculate g of solute from MM
 g solute 
molessolute  moles   gsolute
solute
4. Sum masses to get total mass of solution
1000g solvent  g solute  g solution
5. Calculate V of solution from density
 mLsolution   L 
  3    Lsolution 
 gsolution  g


solution  10 mL
6. Calculate moles/V
 molessolute 

M
 Lsolution 
Example Density of an aqueous solution of KOH is 1.43g/mL and the molality is
14.2. What is the molarity?
1. Assume 1 kg solvent
2. Calculate moles of solute
 14.2molesKOH 
 1kgwater  1kgwater   14.2molesKOH


3. Calculate g of solute from MM
. gKOH 
 39.10  16.00  108
14.2molesKOH 
  797.756 gKOH


moleKOH
4. Sum masses to get total mass of solution
g solution  g H2O  g KOH  1000g  797.756g  1797.756g solution
5. Calculate V of solution from density
 1mLKOHsolution 
  1,257.17mL
1797.756gsolution  143
. g KOHsolution 
6. Calculate moles/V
14.2 moles, KOH
moles
M

 11296
.
Lsolution 1257
.
LKOHsolution
Methods of measurement are related but not equivalent
16
14
45
40
12
35
10
30
25
8
20
6
15
4
10
2
5
0
0
0
2
4
6
8
Molarity
10
12
14
16
weight percent
Molaity or (densityx10)
50
Density
Molality
Weight %
“A” students work
(without solutions manual)
~10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th & F 2-3:30 pm
Module #13
Solution Properties
Water/Salt Solutions 1
SOLUTIONS
Any two components A and B
Mixed in any mole fraction
B
Solvent is usually
Considered the “dissolver”
A
Solute is usually
Considered the “dissolved”
A mixture is usually plotted as variation in mole fractions
nB
 B
n A  nB .
nA
nB
nB  nB
 A  B 


1
n A  nB . n A  nB . n A  nB .
nA
 A
n A  nB .
1
mole fraction
0.8
0.6
0.4
0.2
0
0
Pure B
0.2
0.4
0.6
moles of A
mixture
0.8
1
Pure A
SOLUTIONS
solvent
solute
water
2
2
MgSO4 ,s 
 Mg aq
 SO42,aq
H O
 H   912
.
kJ
mol
Any two components A and B
Mixed in any mole fraction
Solvation Diagrammed as an Example of Hess’s Law
2
2
MgSO4 ,s 
 Mg aq
 SO42,aq
H O
 H   912
.
kJ
mol
-1284.9kJ
mole
-1376.1kJ
mole
NH4 NO3( s)  NH4( aq )  NO3( aq )
 H   281
. kJ
http://www.qtp.ufl.edu/~roitberg/pdf/2002_04.pdf
http://www.lsbu.ac.uk/water/magic.html
 H   281
. kJ
Aquated
Ammonium
Aquated
nitrate
NH4 NO3( s)  NH4( aq )  NO3( aq )
water
NH4 NO3( s)  heat 
NH4(aq)

NO3(aq)
 H   281
. kJ
For most salts,
heat will increase solubility
“A” students work
(without solutions manual)
~10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th & F 2-3:30 pm
Module #13
Solution Properties
Organic/Solute
Solutions
SOLUTIONS
solvent
solute
water
2
2
MgSO4 ,s 
 Mg aq
 SO42,aq
H O
organic
 H   912
.
kJ
mol
What trends do you see here?
s
g solute
g solvent
What trends do you see here?
Like Dissolves Like
similar intermolecular forces between
solute-solute
solvent-solvent
imply solute-solvent interaction will be decent
Soluble in water?
Soluble in water?
Context Slide
Solute-Solvent interactions: Vitamin C and Lead
Vitamin C is more soluble in the aqueous
Phase of the body (urine) than other
Vitamins which are stored in fat.
Vitamin C is water soluble
so it is excreted from the
body
the longest you can last
without serious disease
without consuming fresh
vitamin C is 6 months
Context Slide
Sir Franklin
Context Slide
Context Slide
John Hartnell
Context Slide
“A” students work
(without solutions manual)
~10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th & F 2-3:30 pm
Module #13
Solution Properties
Gas/Solvent Solutions
Raoult’s Law, Freezing
Point, Boiling Pt changes
Osmotic Pressure
SOLUTIONS
solvent
solute
Liquid
Liquid Gas Solution
Component B
Consider first case where A
Is a pure substance
Component A
nA
 A 1
n A  nB .
A  Ag
rate escapes
Number of escapes will depend on
1. Number of molecules with energy>>
intermolecular forces
2. Fraction of molecules with that
energy at some the given
temperature
3. The Surface area available for
escape
4. Escapes independent of what gas
is doing
nA
 A 1
n A  nB .
Agg  A
rate sticking
Number of returns will depend on
1. Intermolecular forces in gas
phase greater than collisional
energy
2. Fraction of molecules at with less
energy than intermolecular
forces at that temperature
3. Sticking followed by dropping does
not depend upon the surface area
Liquid Phase
In a container
Prediction 1
1. T, more escapes, more gas phase A
2. T, less sticking (fewer returns)
Predict:
1. T more gas vapor
2. T less sticking of
gases more gas vapor
Ppure liquid  e
Intermolecular
forces
nA
 A 1
n A  nB .
   Hvaporization 


RT


Exponential represents fraction
of molecules at temperature T
with sufficient energy to break
intermolecular forces
(Hvaporization)
Ppure liquid  e
1    Hvaporization 


T
R

Liquid Phase
In a container
11
TT
TT;;ee ; PAo 
Math for the fraction of molecules with sufficient energy
Is on next slide, if desired
From earlier chapter we learned of kinetic energy:
Maxwell’s Distribution
1 2
E k  mv
2
μ
urms 
3kT
σ
gmolecule
RT
1
E k  mass of
2


3kT

molecule 
mass
of
molecule


1
E k   mass of
2

3kT


molecule  
  mass of molecule 
 3kT 
Ek  

 2 
Review
f  x  e
1 x   2
 

2  
EK
2
However, only a fraction of
molecules at a given emperature that
average speed, and therefore,
Average energy
If you assume a normal distribution
for the “bell curve” you can calculate
the fraction of molecules with an
energy above the average value
f ( x)  e

Emolecule
kT
e

Emole
RT
A  Ag
rate escapes
Number of escapes will depend on
1. Number of molecules with energy>>
intermolecular forces
2. Fraction of molecules with that
energy at some the given
temperature
3. The Surface area available for
escape
4. Escapes independent of what gas
is doing
Agg  A
rate sticking
Number of returns will depend on
1. Intermolecular forces in gas
phase greater than collisional
energy
2. Fraction of molecules at with less
energy than intermolecular
forces at that temperature
3. Sticking followed by dropping does
not depend upon the surface area
Liquid Phase
In a container
Prediction 2
1. Surface Area , less escapes, less gas A
2. Surface Area , no impact on rate sticking
SOLUTIONS
solvent
solute
Liquid
Liquid Gas Solution
Component B
One way to diminish
Surface area is to add solute to
Pure substance A
Component A
nA
 A 1
n A  nB .
A mixture is usually plotted as variation in mole fractions
nA
nB
nB  nB
 A  B 


1
n A  nB . n A  nB . n A  nB .
nA
 A
n A  nB .
1
mole fraction
0.8
0.6
0.4
0.2
nB
 B
n A  nB .
0
0
0.2
0.4
0.6
moles of A
Pure B
mixture
0.8
1
Pure A
nA
99 A
 A 
n A  nB .
1B  99 A
10
Within every layer of liquid (10x10)
10
there are 99A and 1B
How has the surface Area available
for escape changed?
New Surface Area   A Original Surface Area
How will vapor pressure change?
We just made a prediction!
A  Ag
rate escapes
Predict:
1. T more gas vapor
2. Surface area  less gas vapor
3. Escapes independent of gas!!
PA   A PAo
Raoult’s Law
PA0
PA
0
Pure A
PA   A PAo
Raoult’s Law
Assumes effect of “B” is to simply block the surface
No interaction of “B” (solute) with “A” (solvent)
Consequence of Raoult’s Law?
Salted water – less escapes at any given temperature
must raise temperature to compensate for less escape area
= boiling point elevation!
What about freezing?
6 guys oriented
Presence
Of salts lessens
Number of
Successful inte
Prediction less successful encounters, to form ice
The number of molecules leaving the ice is unaffected,
therefore, Must decrease temperature
Boiling Point elevation
o
   Hvaporization   TA,bp  TA,bp 
ln1  B   


R

  TA,bp TAo,bp 
1    Hvaporization 
ln P pure liquid  ln P  

T
R

1    Hvaporization 
o
ln PA,b  o 

R
Tb, A 

I hate drivations
  
o
A
If  small
 o

   H vaporization   TA,bp  TA,bp 
B  


2

R

  TAo,bp


 P    Hvaporization   1
1
ln  2   



P
R
T
T
1
2
1
  



 
 PA,bp     Hvaporization   1
1 
ln o   



o
R
T
T
 PA,bp  

  A,bp
A ,bp 

 A  1  B 
nB   n A
nB
 k TAo,bp
nA
nB
n A  nB
nB
     
 kg A 

A
n  mole
   Hvaporization   1
1 
ln1   B   
 o 

R
T

  A,bp TA,bp 
TAo,bp
TA,bp
   Hvaporization  
ln1  B   


R

  TA,bp TAo,bp
TAo,bp TA,bp





H
nB
vaporization
To
 
2

 A,bp
o
nB  n A
 R TA,bp

Clausius-Clapyeron Equation
PA
nA



A
n A  nB
PA0
If numbers
Similar altho
Not the same
A





k
 TAo,bp
 kg A 


 mole A 
molal B  k '  TAo,bp
1
molal B   TAo,bp
k'
 PA   B PAo
Kb molal B   TAo,bp
Boiling Point elevation
Freezing Point depression
Constants for water
Kb molal B   T
o
A,bp
K f molal B   T
o
A,mp
o
C
Kb  0.52
molal
o
C
K f  186
.
molal
SOLUTIONS
solvent
solute
Liquid
Liquid Gas Solution
Component B
Component A
The other extreme is to
nA
  A  1
n A  nB .
When “A” is completely surrounded by “B”
Ain B liquid  Ag
rate escapes
1 Surface area for A
2 Escapes of A from B when
energy is larger than A/B
intermolecular forces
Ag  Ain B liquid
rate sticking
1. Number of possible A/B collisions
(Partial Pressure of A) (PA)
2. Intermolecular forces of A/B in
gas phase greater than collisional
energy
3. Fraction of molecules with
intermolecular energy greater than
collisional energy increases as T
goes down.
4. No dependence on surface area
Liquid Phase of B
In a container
When “A” is completely surrounded by “B”
Predictions
1. Greater PAmore collisions with B,
greater condensation, greater
solution concentration
PA
A 
h
Math (slide after
next)
Cg  k H PA
A/B intermolecular forces
2. Greater T less collisions in whch
intermolecular forces allow
sticking, less solution
concentration
Liquid Phase of B
In a container
Solubility in water
Henry’s Law
Cg  k H PA
Prediction 1
Prediction 2:
As temp goes up, fewer
Gas phase collisions between
A/B allow them to stick,
Less condensation
if n A   n B
Henry’s Law
PA  break A  B energy A
 nA 

PA  k '' 
 mB 
PA  k A
 nA 
1
  k H PA
PA  
k ''
 mB 
 nA 

PA  k 
 n A  nB 
 nA 
PA  k  
 nB 




 mB  
nA


PA  k ' 
 nB 
 mB  
 n B 
 
 nB  

For the math hungry
Molality
When
mb in kg
Cg  kPA
kH = Henry’s constant
kH 
break
nB
A  B energymolal B
P
Raoult’s Law
Henry’s Law
Slope is
A function of
Slope is a function
Of
interactions
interactions
Pure solvent
Cg  k H PA

A
PA   A PAo
Raoult’s Law
Partial Pressure
Everything in between
Is hard to predict
Vapor Pressure
VP
PA=PAo
Context Slide
60
Reflects dioxin-water interactions
Partial Pressure of dioxin
50
Henry’s Law
Viktor Yushchenko
2004, Ukrainian
Opposition candidate
For president,
Poisoned by dioxin
40
P (mm Hg)
Pdioxin
Raoult’s Law
30
Reflects dioxin-dioxin interactions
Surface area for escape
Cl
O
Cl
Cl
O
Cl
Cl
O
Cl
Cl
O
Cl
20
10
0
0.00
Cl
O
Cl
Cl
O
Cl
0.10
0.20
0.30
0.40
0.50
0.60
Mole Fraction Dioxin in Water
n Dioxin
  Dioxin
nwater  n Dioxin .
0.70
0.80
0.90
1.00
Galen, 170
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Justus von
Thomas Graham
Liebig (1803-1873 1805-1869
Ludwig Boltzman
1844-1906
Gilbert N
Lewis
1875-1946
Henri Louis
LeChatlier
1850-1936
Johannes
Bronsted
1879-1947
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Richard AC E
Erlenmeyer
1825-1909
An alchemist
Count Alessandro G
A A Volta, 1747-1827
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Lawrence Henderson
1878-1942
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
Amedeo Avogadro
1756-1856
Rudolph Clausius
1822-1888
Jacobus van’t Hoff
1852-1911
Niels Bohr
1885-1962
John Dalton
1766-1844
William Thompson
Lord Kelvin,
1824-1907
Johannes Rydberg
1854-1919
William Henry
1775-1836
Johann Balmer
1825-1898
J. J. Thomson
1856-1940
Erwin Schodinger Louis de Broglie
1887-1961
(1892-1987)
Fitch Rule G3: Science is Referential
Jean Picard
1620-1682
Jacques Charles
1778-1850
Francois-Marie
Raoult
1830-1901
Heinrich R. Hertz,
1857-1894
Friedrich H. Hund
1896-1997
Daniel Fahrenheit
1686-1737
Max Planck
1858-1947
Rolf Sievert,
1896-1966
Blaise Pascal
1623-1662
Georg Simon Ohm
1789-1854
James Maxwell
1831-1879
Robert Boyle,
1627-1691
Isaac Newton
1643-1727
Michael Faraday
1791-1867
B. P. Emile
Clapeyron
1799-1864
Dmitri Mendeleev
1834-1907
Svante Arrehenius
Walther Nernst
1859-1927
1864-1941
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
Johannes D.
Van der Waals
1837-1923
Marie Curie
1867-1934
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
J. Willard Gibbs
1839-1903
Fritz Haber
1868-1934
Thomas M Lowry
1874-1936
Werner Karl Linus Pauling Louis Harold Gray
1905-1965
Heisenberg 1901-1994
1901-1976
Same concept for
As  Ag
rate escapes
Ag  As
10
rate sticking
10
Sublimation
Solid phase
In a box
Imagine: Two volumes of water
1. is “pure”


nelectrolyte

  0   1
 nwater  nelectrolytes .
2. the second has large salt mole fraction


nelectrolyte

  0.2   2
 nwater  nelectrolytes .
3. There is a barrier
between them
which
is impermeable
to the
electrolytes
4. Surface
area for
water
escape is
less in cube
2
Barrier (memrane)
Only water (not
solute)
Is allowed to pass
5. Escapes 1> Escapes 2
6. Water moves 1 to 2
7. As water in 2 increases
Pressure builds up
against membrane
n B RT

  B RT
V
Osmotic pressure (pressure driving
water from one compartment to
another) is directly related to the
concentration of the solute, B.
1. This provides a convenient way to measure molar masses.
2. Has huge physiological implications
Visualization
Water will move from the pure solvent to the solute
containing container.
Cucumber
Cucumber placed in
high salt solution
Aka =Pickle
Prune
Prune placed in
pure water
Aka = Plume
Context
When not well regulated the extracellular fluid increases and
edema (bursting of cells) results
Context
A Student who did NOT take
Gen Chem
Water motion across cell
membranes is driven by
osmotic pressure related to the salt
concentration gradient
Typical ion concentrations in vertebrates and invertebrates
Ion
Cell (M)
Blood (M)
K+
0.139
0.004
Na+
0.012
0.145
Cl0.004
0.116
HCO30.012
0.029
X0.138
0.009
Mg2+
0.0008
0.0015
Ca2+
<0.0000002
0.0018
Normal horse blood
http://images.google.com/imgres?imgurl=http://www.vetmed.auburn.edu/distance/clinpath/morphol/sa10.jpg&imgrefurl=http://www.vet
med.auburn.edu/distance/clinpath/morphol1/&h=80&w=120&sz=3&hl=en&start=10&tbnid=jyQ0sNBf3dS1HM:&tbnh=59&tbnw=88&pr
ev=/images%3Fq%3Dcrenation%26svnum%3D10%26hl%3Den%26client%3Dfirefox-a%26channel%3Ds%26rls%3Dorg.mozilla:enUS:official%26hs%3Dk0K%26sa%3DGcells
Normal horse
Red blood cells
Normal RBC
Bumpy Horse blood cells are dessicating
= “crenation”
Hemolysis RBC
Rupture of RBC
www.healthenlightenment.com/hemolysis.jpeg
The equations
PA  P  A
0
A
n B RT

  B RT
V
Raoult’s Laws
 PA   B P
o
A
K f molal B   T
o
A,mp
Kb molal   T
o
B
A,bp
Colligative Properties
Depend on Solute Conc.
Cg  kPA
Derived from
Raoult’s Law
Henry’s Law
Depends primarily on solvent
“A” students work
(without solutions manual)
~10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th & F 2-3:30 pm
Module #13
Solution Properties
Example Calculations
Raoult’s Law, Henry’s Law,
Boiling Pt elevation
Freezing Pt depression
EXAMPLE Calculations
Example A solution contains 102 g of sugar, C12H22O11, in 375 g of water.
Calculate the vapor pressure lowering at 25 oC (vp of pure water = 23.76 mm Hg
Example Calculate the concentration of CO2 in a soft drink that is bottled with
a partial pressure of CO2 of 4.0 atm over the liquid at 25oC. The Henry’s law
constant for CO2 in water at this temperature is 3.1x10-2 mol/Latm.
Example The solubility of pure nitrogen in blood at body temperature, 37 oC,
And one atmosphere nitrogen is 6.2x10-4 M. If a diver breathes air (N2 = 0.78) at a
depth where the total pressure is 2.5 atm, calculate the concentration of nitrogen in
his blood.
Example: You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to your car
radiator, which contains 4450 g of water. What are the boiling and freezing
points of the solution? Kb of water is 0.512 (oC/m) and Kf of water is 1.86 (oC/m).
Osmotic Pressure Example Calculation:
What is the osmotic pressure related to a 0.10 M CaCl2 solution?
Example 1 A solution contains 102 g of sugar, C12H22O11, in 375 g of water.
Calculate the vapor pressure lowering of water at 25 oC (vapor pressure of pure
water = 23.76 mm Hg
What do we know?
sugar in water (sugar= solute; water= solvent)
What equations might apply?
Vapor pressure lowering
 PA   B P
o
A
What information is irrelevant? (Red herring)
none.
Example 1 A solution contains 102 g of sugar, C12H22O11, in 375 g of water.
Calculate the vapor pressure lowering of water at 25 oC (vapor pressure of pure
water = 23.76 mm Hg
o
2 P1
 solute
0.014123.76mmHg  0.335mmHg
 P
Molar mass:
molessolute

molessolvent  molessolute
 sugar 
144
22
176
total
342
 1molsugar 
102 gsugar 

 342.30 gsugar 
 1molsugar 
 1molwater 


  102 gsugar 
18
.
02
g
342
.
30
g

water 
sugar 




375gwater 
 sugar 
12C = 12x12=
22H =
11O = 11x16 =

0.298molsugar
20.8molwater  0.298molsugar

 0.0141
=
Example 2 Calculate the concentration of CO2 in a soft drink that is bottled
with a partial pressure of CO2 of 4.0 atm over the liquid at 25oC. The Henry’s
law constant for CO2 in water at this temperature is 3.1x10-2 mol/Latm.
Solute?
Solvent?
Equation?
Red Herrings?
Cg  kPg
mol 
mol

2
Cg   31
. x10
.
  4.0atm  0124

L  atm 
L
Sig figs
Example 3The solubility of pure nitrogen in blood at body temperature, 37 oC,
and one atmosphere of nitrogen is 6.2x10-4 M. If a diver breathes air (N2 = 0.78) at a
depth where the total pressure is 2.5 atm, calculate the concentration of nitrogen in
his blood.
Solute?
Solvent?
Equation?
Red Herrings?
N2
Blood
Cg  k H Pg
Example 3The solubility of pure nitrogen in blood at body temperature, 37 oC,
and one atmosphere nitrogen is 6.2x10-4 M. If a diver breathes air ( N2 = 0.78) at a
depth where the total pressure is 2.5 atm, calculate the concentration of nitrogen in
his blood.
Unknown, how can we
Get it?
Cg  k H Pg

4 M 
3


Cg   6.2 x10
195
.
atm

1209
.
x
10
M

atm 

6.2 x10 M  k 1atm
4
3
Cg  12
. x10 M
6.2 x10  4 M
4 M
k
 6.2 x10
1atm
atm
Cg  kPg
Unknown how can we get it?
Dalton’s Partial Pressure Law!
PN2  X N2 Ptotal
PN 2   0.78 2.5atm  195
. atm
Example 4: You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to your car
radiator, which contains 4450 g of water. What are the boiling and freezing
points of the solution? Kb of water is 0.512 (oC/m) and Kf of water is 1.86 (oC/m).
Tb  kb  molality
o

C 
 Tb   0.512
 molality
molality 

molality 
mol solute
kg solvent
Tf  k f  molality
o

C 
 Tf   186
.
 molality
molality 

o

C 
 Tf   186
.
 3.62mC2 H6 O2  6.73o C
molality 

3
 10
O2O 
g33g  1 molC
2 H 6H

11molC
10
2 H
6 O
2



molC
10
g

. 100
kgC
H
O


100

2
6  2 
2
6H 2O 

.
kgC

. kgC2 2 H6 6 O
 1kg
62
.07.07
gCgC
O2O
100
2 21kg
  62
  161
. molC2 H 6 O2
2 H 6H
2
6
1
kg
62
.
07
gC
H
O2 2 




 3.62m
molality

2
6
molality
molality
4.450kg
 1kg1kg
 
kg
solvent
4450
gHgH
2 OO 3  
4450
2 10 g3
 10 g 
o

C 
 Tb   0.512
. oC
 3.62mC2 H6 O2  185
molality 

The radiator will boil at 101.85 oC
The radiator will freeze at 32-6.73 oC
=25.27 oC
Osmotic Pressure Example Calculation:
What is the osmotic pressure related to a 0.10 M CaCl2 solution?
n B RT

  B RT
V
mol
mol
LLatm
atm


RT  33010
298
  BBRT
010
..
00.0821
.0821
298KK  7.2atm






LL 
mol
mol KK

Where did this come from?
3 moles solute
2
CaCl2  Caaq
 2Claq
strong electrolye complete
In order to control this pressure the kidney
regulates the amount of Ca2+, Na+ and other ions in the
blood
Lead can cause significant
Blood pressure problems by interfering
Kidney regulation of ions
Context Slide: Preparation for your papers
Reference Data for Risk Assessment
KOW 
Cg  kPg
g solute / Loc tan ol
g solute / Lwater
Your paper
for a compound of interest
s
g solute
g H2O
find, listing source of data
s
Kow
KH
octanol
H3C
Set a reference state to compare all chemicals
solubility in water (s)
solubility between water and octanol (Kow)
KH
OH
Why would this
Be important?
Hint, think about
Vit. C and Vit A
N.Y. Times, April 3, 2007
“A” students work
(without solutions manual)
~10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th & F 2-3:30 pm
Module #13
Solution Properties
Solid/solid solutions
As an example of importance
Of freezing point depression
SOLUTIONS
solvent
solute
B
copper
A
tin
Context Slide
A Different Kind of Phase Diagram
Liquid
Temp
Solid
nB
 B
n A  nB .
nA
 A
100%A
n A  nB .
100%B
Same idea of mole fraction
Bronze phase diagram
as Tin is added to Copper
Context Slide
Liquid Phase
Solid
+
Liquid
Solid phases
Bronze phase diagram
as Tin is added to Copper
2. Grind up and
mix tin to copper
and you can lower
the melting point
Context Slide
Liquid Phase
1. Temperature
Range of early
“campfires”
3. Bonus point:
final material
harder than
copper!!
4. Many early economics
Of societies
Collapsed as fuel
(forests) ran out
5. England’s interest in controlling Ireland
Was initially over wood supply
Solid
+
Liquid
These represent
various unit cell
structures (face
centered cubic, etc.)
adopted
Egyptian Glass how did they do it? Context Slide
Viscosity = measures flow
low viscosity = easy flow
1300B.C.
Campfire Temps
Additives to
The quartz
(SiO2) allow
The glass to
Melt (becomes
Less viscous)
At campfire
Temperatures.
Context
Sugar of Lead =Pb(CH3COO)2
Pb2+





 O
H


 


2 O  C  C  H
 



H





Where does the negative charge reside?
Is this acetate anion q/r large or small?
Is this acetate anion charge dense or not?
Will acetate “hold on” to Pb2+ or will water win?
Context
Pb2+
Renal biopsy
26 year old
Ship paint stripper
Context
Failure of kidneys leads to deposition of
Excess uric acid to joints = Gout
Immersion of the body Up to head creates a
Pressure gradient in the blood volume towards the
head; Body compensates with increased urine to
lower Blood volume to brain; Increased urination
leads to increased loss Of electrolytes (lead)
Context
Initial symptoms are hallucinations
creepy bugs, etc.
Tetraethyl lead poisoning is highly acute leading very rapidly to edema
Why and what does this have to do with osmotic pressure?
Context
PbCH3CH2  4
Tetraethyl lead has what kind of structure?
Hint: Valence Shell Electron Pair Repulsion
model
Valence electrons of atoms
5(4)
20
single bonds 2(4) 8
remainder to Cs
12
VSEPR = AX4=tetrahedral
Tetrahedral arrangement is
important because the lead
is buried beneath the organic
CH3CH2 chains - making it
lipid soluble.

C 


 C  Pb  C 


C 

Floppy CH3CH2
chain buries Pb
Context
PbCH3CH2  4


 C  Pb  C 


Bond strength
- is the Pb-C bond as
strong as a true fully
covalent C-C bond?
Bond enthalpies
C-C 368
Pb-C 206.7kJ/mol
Carbon gets
the electron

C 
C 

Floppy CH3CH2
chain buries Pb
NO!!!
PbCH3CH2  4  PbCH3CH2  3  CH3CH2
PbCH3CH2  4  PbCH3CH2 




3
  CH3CH2

Chloride ion gets

Pb
CH
CH

Cl
 Pb CH3CH2
3
3
2
buried in the
carbon chain center, making it lipid soluble
 Cl
3
Context
Cl
PV= energy
Triethyllead carries chloride ions across the cell membrane
disrupting the energy cycle of the cell and the osmotic
pressures - ultimately results in cell death by “popping” (edema)
Context
Typical ion concentrations in vertebrates and invertebrates
Ion
Cell (M)
Blood (M)
K+
0.139
0.004
Na+
0.012
0.145
Cl0.004
0.116
HCO30.012
0.029
X0.138
0.009
Mg2+
0.0008
0.0015
Ca2+
<0.0000002
0.0018
“A” students work
(without solutions manual)
~10 problems/night.
Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th & F 2-3:30 pm
Module #13
Solution Properties
END