Chapter 6
What is Henry’s law??
pi= p*iL Xil
pi / Xil = p*iL
for non ideal solutions (low solubility)
pi= i Xil p*i
_
dividing by V l or the molar volume of the
mixture (sometimes called Vmix)
X/Vl = Concentration= Ci
_
pi / V l = i Ci p*i
_
pi / Ci = i p*i V l = const ??? =
KiHl
;
If air – water K iH
Is the product of i, p*i Vmix a constant??
Is it different for different compounds??
Does it vary with temperature???
Does it change with concentration?
1
Does it change with salt or ionic content?
How do we measure it?
Chatper 6 Henry’s Law
K iH
pi

Ciw
traditionally
K iH 
K iaw
K iaw
atmi
1

atm
liters
mole
moles liters 1w
Cia

 (dimensionless Henry’s law const.)
Ciw
Cia K iH
Cia


p
pi
i
K iH
Cia K iH
K iH


n / V RT
RT
2
If we go to the Appendix (p.1200, new book) and look at
for Henry’s law values for air-water, we see -log Kiaw ;,
p*iL and Ciwsat are referenced to their states.
how are these -log Kiaw values computed?
Ideally, since KiH= pi*/ Ciwsat
If we go to a unit-less form, Kiaw, where Kiaw = KiH/RT
So, log Kiaw= log { pi*/RT } –log Ciwsat
For anthracene the Appendix has the following data:
log pi*= -3.01(Pa) -log Ciwsat = 6.60 -log Kiaw =2.8
1atm = 101,308 pascals
so 1st we need to get - log pi*in atm
10-3.01 Pa/{ 101335 Pa/atm} = 9.646 x10-9 atm
to change pi* in atm. into Cair; pi*V= nRT; Cair = pi*/RT
R = 0.082 L atm./mole ; T = 298 K ; this gives
Cair=3.94x10-10 moles/literair ; log pi*/RT= -9.40
-logKiaw= -log { pi*/RT } +log Ciwsat
-logKiaw = 9.40 -6.60= 2.80 (and this is the book value)
3
The old book is a lot cleaner; It gives -log poL and -log
pos directly in atmospheres
and –log Cwsat , -log Cssat are in moles/liter and –log KH
in liters atm/mol
So to get log KH in the appendix (p 621)of the old book
for anthracene in (old book, p. 621)
logKH = log pLo – log Cwsat or
logKH = log pso – log Cssat
for a liquid anthracene
log KH = -6.11+ 4.48= -1.63
for a sold anthracene
log KH = -8.1+6.46= -1.64
KH/RT = Kiaw (in new book);
and -log Kiaw= -3.03
4
going back to Henry’s law
K iH
pi

Ciw
As Henry’s law values increase there is a tendency for
higher gas phase concentrations over water i.e.
partitioning is toward air
for high vapor pressure compounds the fugacity in the
gas phase is high
fi = i Xifi* pure liquid
(fi*pure liquid = p*i pure liquid)
High activity(i) coefs. favor partitioning to the gas phase
i.e. Lower KiH and lower ‘s favor the liquid phase. Polar
compounds?
5
Figure 6.2 page 111 (old book)
6
Wash out ratios or W and how fast does the
atmosphere clean up during a rain
Usually defined as the conc. in rain/conc. In air
W = Ciw/Cia = 1/Kiaw
W x Cia = conc in the rain, Ciw , with units of moles/ cc water
or Ciw in units of moles i /cc = moles i/g H20
The rain has an intensity, I with units of
grams of rain sec-1 cm-2
so now
I x Ciw = g rain sec-1 cm-2 x moles i/g H20
Since W = Ciw/Cia = 1/Kiaw
I x 1/Kiaw x Cia = moles of i from the atmosphere hitting
the surface of the earth in the rain per sec-1 cm-2
And this is a flux
7
We will learn in the old book Chapter 10,
Flux / (conc x depth ) = 1st order rate constant in
C = Co e-kt
So if you know the rain intensity, Kiaw and the height of the
atmosphere, you can estimate how fast the atmosphere
will “clean” up with a given rain intensity???
_______________
Flux = I x 1/Kiaw x Cia = moles of i from the atmosphere
hitting the surface of the earth per sec-1 cm-2
If the mixing height of the atmosphere is 300 m
and we have a rain that gives an 1” of water in 2
hours
I = 2.5g cm-2 /(2x60x60 sec) = 3.47x10-4 g cm-2 sec-1
Kiaw phenol = 2x10-5
krate constat = I x 1/Kiaw x Cia / (Cia x30,000 cm)
in units of 1/sec
8
Concentration effects on KiH
Ciw = Xi / Vw
Vw = molar vol. H2O
pi  iw  iw pl*
K iH 

  iw pl*Vw
Ciw  iw / Vw
Under dilute conditions KiH is directly proportional to
the:
 activity coef.
 saturated vapor pressure
 molar volume of water
9
What is the effect of concentration on KiH?
P*iW
water
organic
at saturation the vapor pressure pi = p*iw
pi = i Xi p*i pure liquid
sat
1
p
1
sat
i
X iw  sat *  sat
 iw pi  iw
sat
K iH

p sat
i
sat *


pl Vw
sat
w
C iw
The question becomes how does KiHsat differ from KiH ?
10
If the activity coef. changes with increases in
concentration of Ciw then KHsat will change?
Why?
The old book suggests from benzene partitioning data,
that little difference may exist between KiHsat and KiH
for benzene K’iaw = (Cair/Ciw) a difference of <4% was
observed between saturated and dilute water
solutions….
This means that KiH can sometimes be approx. from
KiHsat and estimated from
sat
K iH

p sat
i
sat
C iw
Example
If the Ciwsat for chlorobenzene = 4.3x10-3 mol/L at 25oC
and p*iL = 1.6x10-2 atm what is the KiH
11
K iH K iH 
sat
p sat
iL
sat
C iw
1.6 x10  2 atm

 3.6atm L / mol
3
4.3 x10 mol / L
K iH
3.6atm L mol 1
K 

 0.15

1

1
iaw
RT 0.082 atm L mol K x 298 K
A simple way of changing iw into iwsat (this does
not always work)
log  i 
sat
K iH

log  isat
(1  xisat ) 2
p sat
iw
sat *


piLVw
sat
w
Ciw
for infinitely dilute solutions
KiH
pi  iw iw p *iL
*


  iw piL
Vw
Ciw iw / Vw
12
Comparison of iw and iwsat


iw -logCiwsat Ciw sat
(Tab 5.2) (p618)
mol/L
iwsat
1/(CsatVmix)
(old book)
benzene
toluene
chlorobenz
hexCl-benz
octanol
2400
12000
19000
9.8E+8
37000
1.64
2.25
2.35
5.56
2.35
0.0229
0.0056
0.00447
2.75E-6
0.00447
2425
9879
12437
2.0E+7
18656
Why are iw values sometimes greater than iwsat?
13
Effect of Temperature
*
ln p i L
vap Hi 1

 const
R
T
by analogy
sat
ln x iw
sat
Ciw
H Eiw 1

 const
R T
sat
x iw

Vmix
so substituting
excess heat of solution
ln Cwsat
K iH
sat
ln KiH
H Eiw 1

 const ( Vmix )
R T
PIl*
 sat
CIw
E
VAP Hi  Hiw

 const H
RT
14
page 115, Table 6.1  vapHi- HEiW = awHiHHenry
15
Figure 6.3 page 116 (old book)
17
What are the effects of salts?
in Chapter 5 the relationship between a
saturated solution in water vs. sea water is
discussed
(Setschenow, 1889)
sat 
 Ciw
log  sat K is [ salt ]tot
Ciw ,salt 
let’s say we want to calculate the equilibrium
distribution of anthracene in sea water,
ie KiH w,salt
if we transform Setschenow’s equation
s
sat
sat
log C iw

log
C

K
, salt
iw
i [ salt ]tot
K i s [ salt ] tot
sat
sat
C iw ,salt Ciw  10
the Henry’s law for salt water is
p*iL
*
piL
K iH ,w , salt  sat

 K i s [ salt ]
sat
Ciw , salt Ciw
10
 K i s [ salt ]
sat
K iH 10
18
for anthracene Kis = 0.3, assume [salt] = 0.5 M
and KiH = 0.078 atm L mol-1
so KIH,w,salt= 0.078x10(0.3)x(0.5)=0.11 atm L mol-1
19
Table 6.2 p 117 (old book)
21
Estimating Henry’s Law values
Hine and Mookerjee (1975)
Log Kiaw =nj x functional groupi
OH
for phenol
there are
(old book)
(new book)
p 206
Table 6.2
6 aromatic carbons at: -.33/carbon
5 aromatic C-H groups, at: 0.21/group
and one C-O group at: 0.74
and one OH group at: -3.21
-0.264
+0.154
-0.596 (C-OH)
-3.232
(old book)
log K’H = 6x(-.33)+5(.21)+0.74+(-3.21) = -3.40
(New book) log Kiaw= -4.64
K’H = 0.0004 ; new book Kiaw= 0.000023
from p*iL / Csatw= 0.00041
22
23
Example Problem: Consider a well sealed flask
with 100 ml of H2O and 900 ml air. At
equilibrium estimate the amount of
chlorobenzene in the air and in the water if the
sum (total) in both phases is 10 g.
fw = the fraction in the water phase
fw = chlorobenzene mass in water/total mass
fw 
C w Vw
1
1


C w Vw  Ca Va 1 Ca Va 1 K  Va
H
Vw
C w Vw
Using the Hine and Mookerjee
Cl
K’H = Kiaw= 0.1622
fw = 1/{(1+0.1662)900/100}=0.41
the concentration in the aqueous phase Cw is
Cw = fw Mtot / Vw
/
C = 0.59x10g /0.9 L =
Cw = 0.41x10g 0.1L = 41 g/Lwater
a
6.6g/Lair
24
Experimental Measurements
1.
air
toluene
McAuliffe (1971)
fract in H2O =
CiwVwv
CiaVg  CiwVwv
Vwv = vol of water
for dilute systems
Kiaw= Cia/Ciw = Dg,w( a gas/water part. coef.)
fract in H2O=
Vwv
K iawVg  Vwv
each time we take a step
n
Cia,n= (fact in H2O) Ciw,o Kiaw
taking the logs of both sides and substituting for
fract in H2O and remembering that Kiaw=Dgw
25
log Cia,n  n log
Vwv
+log (Ciw oDgw)
K iaw Vg  Vwv
Figure 6.4 page 119 (old book)
26
2. Mackay and co-workers experimental KH
technique using a stripping apparatus
Cow initially
bubbles
IF we take this as a CSTR the conc. of Cw
some time, t, after we start the bubbles is
Cw = Cow e-kt
if we were just flowing in clean water instead of
bubbles into some volume of water Vw
Cw = Cow e-f/Vw t
Cw
time
to take into account the gas that is stripping,
f, the flow of water is replaced with
Kiaw x flow rate bubbles
*
Cw = Cow e- Kiaw Vg/Vw t
27
Table 6.3, p120
29
An acid Rain example
Atmospheric acidity of “pure” rain
CO2 +H2O --> CO2H2O
(1)
for reaction 1, KH(CO2) = pCO2 / CO2H2O
CO2H2O dissociates in water
CO2H2O <---> + H+
(2)
the equilibrium const K2 for this reaction is:
[HCO3-] [H+]
K2 =
[HCO3-] [H+] KHCO2
=
CO2H2O
pCO2
bicarbonate reacts to
HCO3- < --> CO3-2+H+
[CO3-2] [H+]
K3 =
(3)
[CO3-2] [H+] KH(CO2)
=
HCO3-
K2 pCO2
we now have expressions for each carbon form
30
When we add up [CO3-2] + HCO3-]+ [CO2H2O] and set this
to [CO2]T
pCO 2 
K1 K1 K 2 
[CO 2 ]T 
1     2 
KHCO 2  [H ] [H ] 
An additional condition of ions in solution is that there be
electrical neutrality, ie.
[H+] = [OH-]+[HCO3-]+2[CO3-2]
H2O  H  OH
0)
[HCO3-] [H+] KH(CO2)
K2 =
2)
pCO2
[CO3-2] [H+] KH(CO2)
K3 =
3)
K2 pCO2
If we substitute for each of the ions in the electro-neutrality
equation
31
[K W ] KH( CO 2)K1pCO 2 2KH( CO 2 )K1K 2 pCO 2
[H ] 


+
+
[H ]
[H ]
[H+ ]2
+
AT a given temperature, KH(CO2) K1,K2 Kw are known. For
pCO2 = 330 ppm, it can be shown that at 283oK the pH will be
~5.6. This is often the value cited for “pure” rain water.
The above equation can be numerically solved in most
spread sheets!, by moving the [H+] on the left to the right
side of the equation.
32
Using fugacities to model environmental
systems (Donald Mackay ES&T, 1979)
Consider the phase equilibrium of five environmental
compartments. Is it possible to tell where an environmental
pollutant will concentrate?
A
B
C
E
D
where A= air, B= lake, C= Soil, D= Sediment, E= biota and
suspended solids
When a system is at equilibrium the escaping tendencies in
each phase are equal
fA = fB = fC = fD = fE
33
For Example: oxygen in water at 0.3 mol/m3 and in air at
8mol/m3 exert the same escaping tendency of 0.2 atm and are
thus in equilibrium with the same fugacity.
1. Fugacitys are linearly related to conc.
oxygen in water at 0.03 mol/m3 exerts
a fugacity of one tenth the fugacity of 0.3mol/m3.
fA = fB = fC = fD = fE
Fugacities can be translated into concentrations
fi Zi = C
where Z is called a fugacity capacity value
-----------------------------------------------------3. the mass Mtotal = Ci Vi = fi Zi Vi
if the system is at equilibrium
Mtotal = fi  Zi Vi
Mi = fi Zi VI
4. Calculating Z values
Zi fi = Ci;
Zi= C/f
34
In air f is equal to the partial pressure,pi
piV = nRT, pi = Cair RT, so
Ziair = 1/RT
at 298K , RT= 0.082 liter atm K-1 mol-1x298K
RT= 0.025 m3 atm mol-1
-------------------------------------------------------------------------------------------------
In water pi = KiH Ciw
pi = KiH Ziw fiw
and Ciw = Z fiw
Ziw = pi /{fw KH}= 1/KH
We will use a representative value of
KiH= 1x10-4 m3 atm mol-1
35
On soils, sediment, and suspended solids
Cwi + S ----> Cis
CiS
K iwS  C xS ;
iW
Cis = KisxCiwxS
Cis =Zi sp x fis and Ciw = pi /KiH
Zi sp = KiwS x 1/KiH x pi x S/fis = Ki sp x S/ KiH
For suspended solids at 1,000 mg/m3 and
a Ki sp of 10-4 m3/mg, Zsp= 103 mol atm-1 m-3
For sediment and soils at 2x109 mg/m3 and
a Kiws of 5x10-5 m3/mg, Zs,s= 109 mol atm-1 m-3
For Aquatic Biota
ZB = B  y  Kiow/KiH
where B is the volB/vol H20= 5x10-6 m3/m3,
y=octanol fract. of B = 0.2, Kiow=105; ZB=104
--------------------------------------------------------------------------------------------------
36
Let’s look at the Equilibrium Distribution of a toxic compound
with an atmospheric concentration of 4 x 10-10 mol/m3.(fi x Zi = C
and Mi = fi Zi Vi)
Z
air
water
s solids
Sed
Soil
Aq biota
40
104
103
109
109
104
Vol
fi
M
(m3)
(atm)
(moles)
1010
106
106
104
105
106
10-11
10-11
10-11
10-11
10-11
10-11
4
10-1
10-2
102
103
10-1
%
0.35
0.01
0.001
9.1
90.5
0.01
g/m3.
10-5
0.01
0.05
0.5
0.2
37