BAHAN
ORGANIK
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Tipe-tipe Substansi (Material) Humik :
1.
Fulvic acid (FA): FA is the fraction of HM that is soluble in
aqueous solutions at all pH values. This due to the fact that it
contains both acidic and basic functional groups.
2. Humic acid (HA): HA is the fraction of HM that is insoluble
under acidic pH conditions (pH ≤ 2) but soluble at pH > 2. This
due to the fact HA contains only acidic groups.
3. Humin material (Hu): Hu is the fraction of HM that is insoluble
at all pH values. This due to the fact that Hu dose not contain
any type of surface functional groups.
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Pembentukan substansi Humik
Humic material is formed via complex sequences of only
partially understood reactions. There are two mechanisms that
can explain the formation of HM:
Degradative theory:
This theory assume that after death of plant, the carbohydrates and
proteins will be degraded and lost by the action of microbial agents. The
other remaining refractory biopolymers (like lignin, paraffinic
compounds, melanins, and cutin) are transformed to produce Humin. The
degradative mechanism can be presented in the following scheme:
Plant Material → Humin → Humic acid → Fulvic cid →
Small molecules
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Pembentukan substansi Humik
Polymerization theory:
This theory assumes that plant biopolymers are initially degraded into
small molecules. The small organic molecules re–polymerized to form
humic substances. As a result of these complex re-polymerization
processes, fulvic acid produces then humic acid and finally humin
material. This mechanism is the reverse of the mechanism of
degradative theory.
The polymerization mechanism can be presented in the following
scheme:
Plant Material → Small molecules → Fulvic cid → Humic
acid → Humin
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Pembentukan substansi Humik
In wet-sediments and in aquatic environments the
formation of HM is occurred via degradative theory
because of the presence of oxygen.
However, the production of HM in soil is occurred via
polymerization mechanism because of the harsh
conditions in the soil which converts the plant into
small molecules immediately after death.
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Composition and structure of humic materials:
Humic substances obtained from different locations have an identical chemical
composition.
Typical chemical composition of most humic substances:
C%
O%
H N% Ash
%
%
45-60 25-45 4-7 2-5 0.5-5
Humic substances (and organic matter) have a high carbon content (up to 60%). Ash comes
from the inorganic materials that present in most soils.
The relation between humic matter content (or organic matter) and organic carbon content:
The mass of HM can be obtained from mass of organic carbon and vice versa. The relation
is:
mass of organic carbon = 0.6 × mass of humic matter (or organic matter)
Or:
mass of humic matter (or organic matter) = 1.7 × mass of organic carbon
The conversion factor (0.6 or 1.7) is adopted assuming that the mass percent of carbon in any
humic
or organichttps://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
matter is approximately 60%
SUMBER:
Example:
The concentration of dissolved organic carbon (DOC) in a certain natural water
system is 2.3 mg/L. Calculate the concentration of organic matter in this water
system.
Solution:
Concentration of organic matter (in mg/L) = 1.7 × concentration of organic carbon (in
mg/L)
Concentration of DOM = 1.7 × 2.3 mg/L = 3.9 mg/L.
The carbon content of humic substances increases in the trend:
Fulvic acid < Humic acid < Humin
The oxygen content of humic substances increases in the trend:
Fulvic acid > Humic acid > Humin
The chemical analysis made for humic materials showed that they contain Several
surface functional groups. The identification of these chemical groups was carried
out using IR and NMR techniques.
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Chemical functional groups present in HM (or in OM)
Functional group
Chemical structure
Content
(mmol/gHM)
O
carboxyl
C
HM
OH
2-6
Phenolic
MH
Alcoholic
Carbonyl
(ketones and
quinones)
Methoxyl
OH
MH
OH
1-4
1-4
O
MH
C
R
(ketone)
2-6
O (quinone)
HM
MH
OCH3
0.2-1
The chemical structure of HM is not known
exactly, however, the enclosed Figure depicts a
generic structure for a humic substance.
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Bentuk-bentuk Substansi Humik:
Humic substances are found in the aqueous and terrestrial environment
in the following forms:
1. Free HM: Consist of soluble or insoluble forms. If
particle diameter is less than 0.45 μm then the HM is
soluble. If particle diameter is more than 0.45 μm then
the HM is insoluble.
2. Complexed HM: Humic matter that is bonded to metal
cations (Mn+), PO43-, or organic molecules.
3. Surface-bonded HM: HM that is bonded to solid surface
such as clay minerals or Iron and aluminum oxides
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Calculation of the molar concentration of surface functional groups present in humic
matter:
HM contains many functional groups that can make complexation with various organic and
inorganic molecules that present in solution. Because surface functional groups are important
for complexation, it is important to calculate their molar concentration in any given sample of
HM.
Example:
Calculate the concentration of carboxyl groups (in μmol/L) that present in 10 mg/L humic
material in a certain water system.
Solution:
As can be seen in previous Table, the typical concentration of carboxyl groups in HM is in
the range: 2-6 mmol/gHM. The average concentration of carboxyl groups is 4 mmol/gHM.
The concentration of carboxyl groups in μmol/L:
= 10 mgHM/L × 4 mmol/gHM (but mmol/gHM = μmol/mgHM)
= 10 mgHM/L × 4 μmol/mgHM
= 40 μmol/L
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Aqueous Humic Materials as Acidifying Agents.
Free HM are an acidic compounds because of the presence of carboxylic groups
(pKa = 2.5–5) and phenolic groups (pKa = 9–10). In water systems that are lacking
natural buffer (hydrogen carbonate HCO3─), HM can acidify water to pH = 5.5–6.5
and achieve the electrical neutrality for water system.
Example:
A sample of natural water containing 8 mg/L dissolved humic material and the
following ions:
Cation
NH4+
Na+
K+
Mg2+
Ca2+
H+ (pH = 5.88)
Concentration
3.6 μg/L (as N)
75.9 μg/L
50.8 μg/L
0.124 mg/L
0.569 mg/L
1.32 × 10-6 M
Anion
Cl─
NO3─
HCO3─
SO42─
Concentration
0.138 mg/L
7.0 μg/L (as N)
14.4 μg/L (as C)
59.4 μg/L (as S)
Calculate the concentration of carboxylate groups in HM that will achieve
elctroneutrality for the water system.
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Solution:
The total positive and negative charges should be calculated using molar
concentrations
Total positive charge = 44.7 μmol/L
Total negative charge = 9.4 μmol/L
This water system is not neutral, it contains more positive charges.
Excess positive charge = 44.7 – 9.4 = 35.3 μmol/L
This amount of positive charge should be balanced by negative charge to produce a
neutral water system. In fact the only source of negative charges is only HM. At pH
> 5 (pKa of HM), the humic material will be fully deprotonated and carry a net
negative charge. pH of this water system is 5.88 and it is higher than pKa of HM. To
achieve elctroneutrality, the HM should provide 35.4 μmol/L of negative charges.
It is known that the concentration of carboxylic acid (or carboxylate groups)
= 2 – 6 mmol/gHM. The equivalent concentration of carboxylate groups in (μmol L─1)
is 16─48. In fact HM contains enough amount of negative charge to balance the
https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
extra SUMBER:
positive charge
in solution (35.4 μmol/L).
Mechanisms of interaction of organic molecules with hm.
Mechanism 1: Electrostatic Attraction
Interaction between positively charged molecules and negatively
charged HM (electrostatic attraction).
Example for this mechanism is the interaction between HM and atrazine
at pH = 8. At this pH, atrazine is positively charged and HM is
negatively charged.
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Mechanism 2: Hydrogen Bonding.
The hydrogen bonding reactions are possible, involving
oxygen, and nitrogen-containing functional groups of both
HM and organic solute.
Example for this mechanism is the interaction between
carbaryl (an insecticide) and
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Mechanism 3: Salt linkage or salt bridge:
In this mechanism a salt bridge is formed between the negative surface
charge of HM and the negatively charged organic solute. Example on
this mechanism is the interaction between 2,4-dichlorophenoxyacetic
acid (2,4-D) and HM molecules at pH 6-8 . At this pH range, both HM
and 2,4-D are both negatively charged.
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Mechanism 4: Hydrophobic interactions:
The interaction of non-polar molecules with HM is occurred via
hydrophobic – hydrophobic interactions. The non-polar molecules will
interact with the non-polar part (or hydrophobic part) of HM. An
example on this mechanism is the interaction between DDT
(dichlorodiphenyltrichloroethane) with HM molecules .
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Interaction of HM with clay minerals.
There are three mechanisms of interaction between clay minerals and HM molecules:
1.Specific adsorption between Al3+ & Fe2+ that present on clay surface and HM molecule.
2.HM and clay minerals can be bonded via a salt bridge (Mechanism 3). Ca2+ and Al3+ can
work as a salt bridge.
3.Hydrogen binding. Clay minerals contain surface hydroxyl groups (─OH); these polar
groups can make hydrogen bonding with polar groups present in HM molecule.
Specific adsorption: M: Al3+ or Fe3+
Hydrogen bonding
Salt bridge: M: Al3+ or Ca2+
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Metals in the
Hydrosphere
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Introduction:
Humic materials, whether it is dissolved in water or present as part of the solid phase in
soils and sediments, have functional groups that are capable of acting as ligands in
forming complexes with metals. Most metals can interact with HM. The functional
groups available for complexation reactions are presented below.
A possible reaction between Pb2+ ions and a portion of HM is given the following
reaction:
O
O
HM
C
+
Pb2+
MH
O
OH
OH
+ 2H+
C
O
Pb
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Factors that affecting the complexation of HM with metal cations:
a)Nature of the bond formed: Covalent versus Ionic bond:
Alkaline earth metals form a weak covalent bond with the negative sites on the HM.
While, the other heavy metals have large stability constants with HM, so, they form
very stable covalent bonds. For example: Cu2+ and Pb2+ tend to be more strongly
bonded with HM than Ca2+ and Mg2+ ions. This because Cu2+ and Pb2+ are able to
make covalent bonds with HM, while Ca2+ and Mg2+ are not able to make covalent
bonds with HM and prefer to make ionic bonds.
b) pH-value:
At acidic solutions (low pH), H+ will be in high concentrations, and so, H+ ions will
compete metals for active sites on HM and decrease metals complexation.
c) Ionic strength (IS):
IS can be generated from salts like: NaCl, KCl, NaHCO3 and Na2SO4. Ionic strength of
the solution is inversely related to the complexation of metal cations with HM. This can
be attributed into two reasons:1) a competition of (Na+and K+) with metal cations for
the negative active site on HM at high IS; 2) the high concentration of anions (Cl─,
SO42─, and HCO3─) will increase the complexation of these anions with metal cations
and hence
reducehttps://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
their complexation with HM.
SUMBER:
(d). Availability of functional groups:
The maximum complexation between metal cations
and HM functional groups occurs at 1:1 ratio.
This indicates that the complexation process is a
one-step process.
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Complexation reactions of metal cations with HM.
The complexation reaction between the functional groups in HM and dissolved metal
cation (Mn+) can be presented as:
M
n
 HM  M  HM
[ M  HM ]
K f  formation cons tan t 
[ M n  ][ HM ]
The H─HM complex should be stable only at: a) lower ionic strength solution, b)
moderate to high pH values, c) Mn+ prefers to make covalent bond, and d) 1:1
complexation occurs.
The value of Kf is changing with these variables (a-d). Therefore, Kf values always
calculated at fixed conditions of pH, ionic strength, and temperature (operationally
defined). The pH value has a significant effect on Kf values
In most calculations, a conditional formation constant is used (Kf′ ), which is the
formation constant (Kf), that is calculated at certain pH value:
Kf′ = Kf × C
Kf′ : the conditional formation constant.
Kf: the formation constant.
C: constant
which
is calculated at certain pH.
SUMBER:
https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
Conditional formation constants (Kf′) at pH = 5.0 for soluble fulvic acid (as a form
of HM) with some metals.
Mg2+
Ca2+
Mn2+
Co2+
Ni2+
Cu2+
Zn2+
Pb2+
Kf′
1.4×102 1.2×103 5.0×103 1.4×104 1.6×104 1.0×104 4.0×103 1.1×104
Example:
A natural water sample (at pH = 5) containing 85 μg/L (1.45 μmol/L) of Ni2+ ions and
8 mg/L of soluble fulvic acid (a form of HM). Assume that the concentration of
functional groups in fulvic acid that are capable of binding with Ni2+ is 5.0 mmol/g.
Calculate the concentration of complexed nickel and the un-complexed nickel. The
interaction of Ni2+ by the functional groups present in fulvic acid can be imagined as:
Ni2+ + FA → Ni─FA
Kf′ = 1.6×104
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SOLUTION
The concentration of functional groups should be calculated in (μmol/L or mol/L):
Concentration of functional groups in FA = 8.0 mg/L × 5 mmol/g (or 5 μmol/mg)
= 40 μmol/L (concentration of binding sites)
Initial (M)
Change
(M)
net
(M)
Change
(X in M)
Equilibrium
(M)
Ni2+
1.45×10-6
+
FA
40×10-6
─1.45×10-6
─1.45×10-6
zero
3.9×10-5
+X
⇌
+X
3.9×10-5 +X
X
Ni---FA
zero
+1.45×10-6
1.45×10-6
─X
1.45×10-6─X
At equilibrium
Kf'
(1.45  10 6  X )
 1.6  10 
( Assume that X  3.9  10 5 )
5
( X )(3.9  10  X )
4
(1.45  10 6  X )
7
1.6  10 

X

8
.
93

10
M
5
( X )(3.9  10 )
4
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SOLUTION
The concentration of un-complexed Ni2+ = X = 8.93×10-7 mol/L.
The concentration of complexed Ni2+ = cinitial – cequilibrium
= 1.45×10-6 ─ 8.93×10-7 = 5.6×10-7 M
The percentage of un-complexed Ni2+ =
The percentage of complexed Ni2+ =
[ Ni]uncomplexed
[ Ni]initial
[ Ni]complexed
[ Ni]initial
8.93 10 7
2
 10 

10
 61%
6
1.45  10
2
5.6 10 7
2
10 

10
 39%
6
1.45 10
2
Note: The complexed nickel (39%) stays in the environment for a long time because the
complexation with HM keeps Ni2+ ions in solution for a long time.
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Metal complexes with ligands of anthropogenic origin:
There are many complexing agents that have anthropogenic sources:
a)NH3: result from nitrogen-containing organic waste.
b)SO32- and SO42-: result from pulp and paper mills.
c)PO43-: result from detergents industry.
d)CN- : result from gold industry.
e)EDTA: result from paper, detergent industry.
f)NTA (nitrilotriacetic acid, H3T): result from detergent industry.
Last two complexation agents are strong complexing agents for most metals in solution.
A tetrahedral complex that is formed between NTA and Cu2+ ions in solution.
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Environmental Chemistry of Colloids
and Surfaces
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Introduction:
There is a clear connection between the hydrosphere and terrestrial (solid)
environment. This connection contains suspended solids of very small particle
size that can interact with ions in water. These suspended matters (or sediments)
are called colloids and they come from the solid part of the environment.
some of natural colloids present
in natural water.
Colloid
1. SiO2
2. MnO2
3. Fe2O3
4. Al2O3
5. Humic material (humic and
fulvic acid)
6. Montmorillonite (clay type)
7. Kaolinite (clay type)
sediment/water system and soil/water system.
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the size-range distribution of some components of natural water system including
colloidal materials.
1.
2.
3.
The diameter of colloidal particles falls in the range 0.01 μm to10 μm.
Particle size less than 0.45 μm is considered to be soluble in water (true solution)
Particle
size https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
higher than 0.45 μm is considered to be insoluble(i.e., colloidal solution) .
SUMBER:
Why do colloidal materials
adsorb different solutes
(metals and organics) in
solution?
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Adsorption Process.
Adsorption is a physical or a chemical process in which a solute
(adsorbate) is transferred from solution to the adsorbent surface.
Adsorption finished when the active sites (adsorption sites) filled
with adsorbate molecules.
Chemical or physical
bond between
adsorbate molecule
and surface site
The adsorption of an adsorbate onto a colloid surface.
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Factors control adsorption process:
Colloidal materials adsorb metals and organics because they have large specific surface
area and contain surface functional groups.
Specific surface area:
Most colloids have a large specific surface area. As surface area increases the amount
of adsorbed ions increase.
The specific surface area of some colloids.
Colloid
Specific surface area
(m2/g)
Montmorillonite
5-20
Kaolinite
700-800
Humic and fulvic aid
700-1000
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Example:
Calculate the particle diameter (in m) for Montmorillonite assuming that the density of
this material is 1.7 g/ml.
Solution:
The average specific surface area of Montmorillonite is 12.5 m2/g. Particle diameter
can be calculated from the following relation:
Specific surface area
(m2/g)
12.5 m2/g =
6
= dp  density
6
dp  1.7  10 6 g m 3
dp (particle diameter) = 3×10─7 m (0.3 μm)
Surface charge:
Most colloidal substances have surface functional groups. These groups played an
important role for attracting the adsorbate molecules from water. Adsorption of
adsorbate molecules on colloids decreases the concentration of adsorbate in solution.
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Types of Adsorption Processes:
1.Electrostatic Adsorption:
Colloidal materials can acquire surface charge in solution. The nature of this surface
charge depends on the pH of the solution in which they present (i.e., the pH of
surrounding solution). One can determine the net surface charge (positive, negative or
neutral) on a colloid surface by comparing the solution pH and the pHo of the colloid.
pHo (pH at which the net surface charge of colloidal material is zero) is constant and
can be determined experimentally for any colloidal material. If solution pH > pHo then
the colloid surface charge is negative and if pH < pHo then the net surface charge is
positive and if pH = pHo then the net surface charge is zero (neutral surface).
pHo values of various natural colloids
Colloid
SiO2
MnO2
Fe2O3 (Hydrated)
pH0
2.0
4.0
7.0
Colloid
Fe2O3 (Haematite)
pH0
8.5
Al2O3 (Hydrated)
7.0
Humic material
4-5
Fe2O3 (Goethite)
7.5
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Humic material has a variably charged surface. Negative surface charge is due to
deprotonation of carboxyl groups, and positive charge is due to amino groups
protonation. Generally speaking, the pH of natural water is about 8.7; therefore, the
majority of natural colloids have a net negative surface charge in solution. This
negative charge is balanced by the positive ions (like Na+, K+) present in water system.
Based on that it can be imagined the colloidal particles in natural water as:
Na+-O
O- Na+
Na+-O
O-Na+
Colloid
particle
O-
-O
Na+
Na+
-O
Na
+
O- +
Na
The ion-exchange reactions (electrostatic adsorption) of colloidal materials with
cations is one of the suggested mechanism in solution and can be presented as:
Colloid-O─Na+ + K+ ⇌ Colloid-O–K+ + Na+
In soil and sediments, the importance of cations for ion-exchange with colloidal
materials is:
2+ > Mg2+https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
SUMBER:
Ca
> K+ > Na+
H+ competition:
In lower pH solutions, H+ can displace the cations that present on the colloid surface as
following:
Colloid-O─Na+ + H+ ⇌ Colloid-O–H+ + Na+
This is known as a competition of H+ with the cations for the binding sites of humic
substances.
The number of negative exchange sites, which equal to the number of positive charge,
on a colloid material is the called cation exchange capacity (CEC).
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2. Specific adsorption:
In this type of adsorption, a strong interaction between the colloid and solute is
established. This type of adsorption occurs at pH = pHo of the colloid. The adsorption
of organic acids (like fatty acid) by colloidal iron oxides surface is a known example for
specific adsorption:
Fe
OH
O
Fe
OH
C
OH
R
+ H2O
O
Fe
OH
Fe
O
C
R
Metal cations are also able to form specific bonds with iron-oxide:
Fe
OH
Fe
O
2+
+ Zn
Fe
OH
Zn + 2H+
Fe
O
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The electrical double layer:
When a colloid material is added to water, an electrical potential is developed around
the colloid surface and this occurs due to the balancing of the negative charge on the
colloid with positive charge that come from surrounding solution.
The electrical potential is highest at the
colloid surface and decrease to zero at the
bulk of the solution. The colloidal system
is stabilized in solution because the small
charged particles repel each other in
solution.
Electrical
double layer.
SUMBER:
https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
Na+-O
Na+-O
Na+-O
O- Na+
+
Na -O
+
Na -O
O-
Na
-O
Na+
O-
Na
Na
Na
Na+-O
O-
-O
+
O- Na+
+
Colloid
particle
Na+
O-
Na -O
O-
+
Colloid
particle
+
O- Na+
Na
O- Na+
Colloid
particle
+
O- Na+
-O
Na+-O
Na+
O-Na+
Na+-O
O-Na+
+
Electrostatic repulsion between colloid particles
These particles are unable to come close to aggregate into settleable particles, and
this due to electrical repulsion. In fact adding strong electrolyte (HCL or NaCl)
will reduce the thickness of the double layer, the surface potential will decrease and
coagulation (aggregation) started.
The process of coagulation is called some times
sedimentation and it is usually occurred when river
colloids encounter with the high-salt sea water.
Electrical double layer in a salty solution.
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Quantitative description of Adsorption:
Adsorption can be described quantitatively by two mathematical equations:
Langmuir equation:
This relation assumes the following hypothesis:
•Number of active site that is capable of reacting with adsorbate molecules is limited.
•No adsorption after filling these sites
•No interaction between adsorbed molecules
Langmuir equation is:
Cs
bCsm

Caq 1  bCaq
Where;
Cs: concentration of adsorbate molecules on colloid surface (mol/g)
Caq: concentration of adsorbate molecules remaining is solution (mol/L)
b: binding constant (L/mol)
Csm: maximum quantity of adsorbate adsorbed on the colloid surface (mol/g)
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The linear form of Langmuir equation is:
1
1
1


Cs Csm bCsmCaq
(2)
A linear plot between (1/Cs versus 1/Caq) will generate a straight line with (1/bCsm) as a
slope and (1/Csm) as an intercept.
Example
Phosphate adsorption by a sedimentary material can be presented by Langmuir
equation. The following data were obtained form phosphate adsorption:
Equilibrium "P" concentration
Sample 1:
Sample 2:
in water Caq (mol/L)
4.0 × 10-7
1.3 × 10-7
in solid (sediment) Cs (mol/g)
2.0 × 10-7
1.0 × 10-7
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Solution:
Applying Langmuir equation to sample 1:
1
1
1


2  10 7 C sm bC sm 4  10 7
Applying Langmuir equation to sample 2:
1
1
1


1  10 7 C sm bC sm1.3  10 7
Solving the above linear equations, b and Csm can be obtained:
b = 2.7×106 L/mol
Csm= 3.8×10-7 mol P/gsediment (or 12.8 μg P/gsediment).
The last number indicate that 12.8 μg of P is adsorbed on one gram sediment.
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The Freundlich Equation:
Adsorption on colloids or soil can be presented by Freundlich empirical equation:
Cs  K F Caq
n
Where:
Cs: concentration of adsorbate molecules on colloid (mol/g).
Caq: concentration of adsorbate molecules remaining is solution (mol/L).
KF: Freundlich constant (L/g).
n: dimensionless number (n is usually less than 1.0).
The linear from of Freundlich equation is:
LogCs = LogKF + nLogCaq
A linear plot between LogCs versus LogCaq will generate a straight line with (n) as a
slope and (LogKF) as an intercept.
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Notes on Freundlich equation:
1.Adsorption of adsorbate on solid surface becomes more difficult as more and
more adsorbate accumulate (i.e. non-linear adsorption).
2. No maximum capacity can be estimated from Freundlich relation. This equation
is valid only at lower concentration ranges.
Example
For a frost soil, it has been reported that KF = 0.0324 L/g, n = 0.82 for Cd metal adsorptio
Calculate Cs if Caq = 4 μg/L.
Solution:
Cs  K F Caq
n
Cs = (0.0324 L/g)(4 μg/L)0.82
Cs = 0.1 μg Cd/g
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Partitioning of small organic solutes between water and soil (or colloid surface):
1. The distribution Coefficient Kd:
At very dilute concentration, n in Freundlich equation can be approximated to unity,
therefore, Freundlich equation becomes:
Cs  K d Caq ; K d 
CS
Caq
Usual units of Kd is (L/kg or mL/g)
Cs: concentration of adsorbate molecules on colloid or soil (mol/kgsoil)
Caq: concentration of adsorbate molecules remaining is water (mol/L)
Notes
If Kd >> 1, then the pollutant is mainly adsorbed on the soil
If Kd << 1, then the pollutant did not adsorb on the soil and remain in water.
The value of Kd depends on: Organic solute itself, the chemical and physical nature of the solid
phase (soil), temperature, ionic strength of the solution. Because these variables are different
from soil to soil, then it is impossible to tabulate values of Kd for pollutants.
In fact, KD value can be correlated to a number of distribution coefficients; KOM, KOC, and KOW.
These coefficients can predict the movement of organic pollutants (like pesticides) and inorganic
pollutants
(like metals)
within soil.
SUMBER:
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2 . Sorption of organic and inorganic pollutants by soil:
The effective parts of the soil that can interact with pollutants are only: organic matters
and clay minerals.
The surface functional groups present in clay and organic matter that can interact with
pollutants:
Clay: (-Si-OH).
Organic matter : (OM-COOH, OM-OH, OM-R-O-R, OM-R-NH2, OM-R-CO-R, and
OM-R-COH).
The type of interactions of various pollutants with clay minerals and organic matters are
mainly: Van der waals, induced dipole-dipole, dipole-dipole, H-bonding forces, and
hydrophopic-hydrophopic interactions.
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2 . Sorption of organic and inorganic pollutants by soil:
Compared to clay minerals, organic matters (OM) are more effective to adsorb
pollutants from solution. This because of the presence of many functional groups
in OM compare to clay minerals.
Most soils contain organic matters (OM). The percentage of OM in soil is in the
range (1–3%). Clay minerals also present in soil in small quantities.
The concentration of a solute in the soil can be presented as:
Cs = fOMCOM + fMMCMM
Where
fOM and fMM are the fractions of organic matter and mineral matter in the soil.
COM: Concentration of solute, or adsorbate, or pollutant in the organic matter
component of the soil or sediment (mol/gOM or mol/KgOM).
CMM: Concentration of solute, or adsorbate, or pollutant in the mineral matter
component of the soil or sediment (mol/gMM or mol/KgMM)
Assuming
that Chttps://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
SUMBER:
MM is so small, then: Cs = fOMCOM
3. Organic Matter-Water Partition Coefficient (KOM) and Organic Carbon-Water
Partition Coefficient (KOC):
The distribution of pollutants in water-soil systems can be described by calculating KOM
and KOC:
Organic Matter-Water Partition Coefficient: K OM  COM
Caq
Organic Matter-Water Partition Coefficient: KOC  COC
Caq
Where;
Caq: Concentration of solute, or adsorbate, or pollutant in water (mol/L).
COM:Concentration of solute, or adsorbate, or pollutant in the organic matter
component of the soil or sediment (mol/gO.M or mol/KgO.M).
COC:Concentration of solute, or adsorbate, or pollutant in the organic carbon
component of the soil or sediment (mol/gO.C or mol/KgO.C).
Note that if KOM and KOC values are greater that 1, then the pollutant is mainly
immobilized (adsorbed) on the surface. If the values of KOM and KOC are less that 1,
then the pollutant is present in solution and has low affinity for soil.
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4. The relation between Kd, KOM, and KOC coefficients:
Kd
Cs

C aq
and
Cs  f OM C OM
then
Kd
but
f OM C OM

C aq
K OM  C OM / C aq
K d  f OM K OM  C s  f OM K OM C aq
K OC  1.7 K OM
SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
Octanol-Water Partition Coefficient (KOW):
The experimental determination of Kd, KOM, and KOC values for pollutants is not
practical because of the variations in the chemical and physical properties of soils.
Based on that, environmental scientists developed a test to predict the movement of
various pollutants in soil without determination of Kd, KOM, and KOC values. n-octanol
was selected because this compound can perfectly function as organic matter that
present in the soil. n-octanol has hydrophilic and hydrophobic components:
Hydrophobic part
OH
Hydrophilic part
KOW value can be used as a guide to predict the movement of pollutants in
soils.
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Octanol-Water Partition Coefficient (KOW):
The distribution of a pollutant between water and octanol can be carried out by adding a
certain amount of the pollutant to a mixture of water and n-octanol. After shaking the
mixture, the equilibrium concentration of the pollutant is determined in each layer. The
n-octanol/water distribution coefficient can be calculated as follows:
K OW 
Cnoc tan ol
Caq
Where:
KOW is the n-octanol/water distribution coefficient (Lwater/Loctanol)
Coctanol: Concentration of solute, or adsorbate, or pollutant in the octanol layer
(mol/Loctanol)
Caq: Concentration of solute, or adsorbate, or pollutant in aqueous solution (water)
(mol/Lwater)
Note
if KOW < 1, then the solute remains in the aqueous phase and has a low molar mass and also has high oxygen
content.
If KOW > 1, then the solute prefer to stay in octanol layer and this also indicates that the pollutant has a high
molar mass and a large carbon to oxygen ratio.
Solutes of lower values of KOW do not adsorb on soil and remain in solution.
SolutesSUMBER:
of higher values
of KOW are highly adsorbed on the soil.
https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt
Solutes of higher molar masses have higher KOW
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The relation between KOW and KOM:
In fact there is a high correlation between KOW and KOM as
shown below:
LogKOM  0.82LogKOW  0.14
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Clay minerals :
Clay minerals are natural type of colloids that distributed throughout
the word. Clay minerals are composed of "aluminosilicate minerals"
with a layered lattice structure. Clay minerals are of terrestrial origin
and carried to water (hydrosphere) in run-off and by winds. Clay
minerals have SiO42– tetrahedral units linked together in a planer
structure by oxygen atoms. There are many types of clay minerals:
kaolinite, halloysite, smectite, vermiculite, and chlorite. Each one of
Material
Molecular formula
these
minerals
has its own CEC capacity. CEC
cm(+)/kg
CEC values
Kaolinite*
Al2Si2O5(OH)4
*
Al2Si2O5(OH)4.2H2O
ofHalloysite
some
natural
clay minerals and some
*
Smectite
(Na,Ca)0,3(Al,Mg)2Si4O10(OH)2 nH2O
colloids.
Vermiculite*
(Mg,Fe,Al)
3(Al,Si)4O10(OH)10.4H2O
Chlorite*
(Fe, Mg, Al)6(Si, Al)4O10(OH)8
Hematite
Fe2O3
Feldspar
NaAlSi3O8
Quartz
SiO2
Organic
See Figure (1) chapter 12
matter
8
8
other
natural
100
125
25
≈4
2
2
200
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Cation Exchange Capacity of Clay Minerals:
The total cation exchange capacity of clays measures its ability to exchange with
cations in solution. Clay minerals can undergo simple ion-exchange reaction with
cations present in solution as following:
+
Na+-O
K
O- Na+
Na+-O
K+-O
O- Na+
O- K+
+ nK+
Clay
particle
O- Na+
-O
+
O- K+
-O
+ nNa+
Clay
particle
O- K+
-O
+
Na
K
Na+-O
K+-O
O-Na+
O-K+
If the amount of released Na+ is determined, then the total cation exchange capacity of
the clay or soil-containing clay can be determined from the following relation:
CEC 
cmol ()
kgclay
Note that K+ is strongly adsorbed compare to Na+ on the clay surface.
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Example:
1.0 gram of a clay sample (in Na-form) was mixed with 100.0 mL of 1.0 M of KNO3
solution. After filtering the colloid clay sample, Na+ ions in the extract was analyzed
using
atomic absorption spectroscopy and found to be 250 ppm. Calculate the CEC for
= 0.1087 cmol(+)
this clay sample in cmol(+)/kgclay.
Solution:
We assume that all Na+ ions released from the clay sample were replaced by K+.
Total amount of Na+ released = 250 mg/L × 0.1 L = 25 mg Na (or 0.025 g Na)
Moles of Na+ =
0.025 g
 1.087  10 3 mol
23.0 g / mol
Mole of positive charge = 1.087× 10-3 mol × 1 (Na+ contains one positive charge)
= 1.087 × 10-3 mol (+) charge
(note that cmol = 10-2 mol or mol = 102 cmol)
= 1.087 × 10-3 mol (+) cmol
= = 0.1087 cmol(+)
10  2 mol
Based on that, we can write:
0.1087 cmol(+) → 1.0g
?
→ 1000.0 g (1.0 kg)
CEC SUMBER:
≈ 109 cmol(+)/kg
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Example:
A sedimentary material containing 8% organic matter, 41%
clay minerals [70% Kaolinite and 30% chlorite]. Calculate the
total CEC (cmol (+)/kg) of this sedimentary material.
Solution:
Using previous table:
CEC = 0.08 × 200 + (0.41) ×(0.70) ×(8) + (0.41) ×(0.3) ×(25)
= 21 cmol (+)/kg
(CEC for organic matter) (CEC for kaolinite)
(CEC for chlorite).
SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt