1. BAHAN ORGANIK SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

2. 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

3. 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 SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

4. 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 SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

5. 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

6. 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%H% N%Ash % 45-6025-454-72-50.5-5 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 organic matter is approximately 60% SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

7. 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

8. Functional groupChemical structureContent (mmol/g HM ) carboxyl 2-6 Phenolic 1-4 Alcoholic1-4 Carbonyl (ketones and quinones) 2-6 Methoxyl0.2-1 Chemical functional groups present in HM (or in OM) The chemical structure of HM is not known exactly, however, the enclosed Figure depicts a generic structure for a humic substance. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

9. 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 (M n+ ), PO 4 3-, or organic molecules. 3.Surface-bonded HM: HM that is bonded to solid surface such as clay minerals or Iron and aluminum oxides SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

10. 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/g HM. The average concentration of carboxyl groups is 4 mmol/g HM. The concentration of carboxyl groups in μmol/L: = 10 mg HM /L × 4 mmol/g HM (but mmol/g HM = μmol/mg HM ) = 10 mg HM /L × 4 μmol/mg HM = 40 μmol/L SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

11. Aqueous Humic Materials as Acidifying Agents. Free HM are an acidic compounds because of the presence of carboxylic groups (pK a = 2.5–5) and phenolic groups (pK a = 9–10). In water systems that are lacking natural buffer (hydrogen carbonate HCO 3 ─ ), 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: CationConcentrationAnionConcentration NH 4 + 3.6 μg/L (as N)Cl ─ 0.138 mg/L Na + 75.9 μg/LNO 3 ─ 7.0 μg/L (as N) K+K+ 50.8 μg/LHCO 3 ─ 14.4 μg/L (as C) Mg 2+ 0.124 mg/LSO 4 2─ 59.4 μg/L (as S) Ca 2+ 0.569 mg/L H + (pH = 5.88)1.32 × 10 -6 M Calculate the concentration of carboxylate groups in HM that will achieve elctroneutrality for the water system. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

12. 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/g HM. 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 extra positive charge in solution (35.4 μmol/L). SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

13. 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

14. 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 SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

15. 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

16. 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

17. Interaction of HM with clay minerals. There are three mechanisms of interaction between clay minerals and HM molecules: 1.Specific adsorption between Al 3+ & Fe 2+ that present on clay surface and HM molecule. 2.HM and clay minerals can be bonded via a salt bridge (Mechanism 3). Ca 2+ and Al 3+ 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: Al 3+ or Fe 3+ Salt bridge: M: Al 3+ or Ca 2+ Hydrogen bonding SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

18. Metals in the Hydrosphere SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

19. 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 Pb 2+ ions and a portion of HM is given the following reaction: SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

20. 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: Cu 2+ and Pb 2+ tend to be more strongly bonded with HM than Ca 2+ and Mg 2+ ions. This because Cu 2+ and Pb 2+ are able to make covalent bonds with HM, while Ca 2+ and Mg 2+ 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, NaHCO 3 and Na 2 SO 4. 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 ─, SO 4 2─, and HCO 3 ─ ) will increase the complexation of these anions with metal cations and hence reduce their complexation with HM. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

21. (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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

22. Complexation reactions of metal cations with HM. The complexation reaction between the functional groups in HM and dissolved metal cation (M n+ ) can be presented as: The H─HM complex should be stable only at: a) lower ionic strength solution, b) moderate to high pH values, c) M n+ prefers to make covalent bond, and d) 1:1 complexation occurs. The value of K f is changing with these variables (a-d). Therefore, K f 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 (K f ′ ), which is the formation constant (K f ), that is calculated at certain pH value: K f ′ = K f × C K f ′ : the conditional formation constant. K f : the formation constant. C: constant which is calculated at certain pH. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

23. Conditional formation constants (K f ′ ) at pH = 5.0 for soluble fulvic acid (as a form of HM) with some metals. Kf′Kf′ Mg 2+ Ca 2+ Mn 2+ Co 2+ Ni 2+ Cu 2+ Zn 2+ Pb 2+ 1.4×10 2 1.2×10 3 5.0×10 3 1.4×10 4 1.6×10 4 1.0×10 4 4.0×10 3 1.1×10 4 Example: A natural water sample (at pH = 5) containing 85 μg/L (1.45 μmol/L) of Ni 2+ 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 Ni 2+ is 5.0 mmol/g. Calculate the concentration of complexed nickel and the un-complexed nickel. The interaction of Ni 2+ by the functional groups present in fulvic acid can be imagined as: Ni 2+ + FA → Ni─FA K f ′ = 1.6×10 4 SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

24. 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) Ni 2+ + FA ⇌ Ni---FA Initial (M) 1.45×10 -6 40×10 -6 zero Change ─1.45×10 -6 ─1.45×10 -6 +1.45×10 -6 (M) net zero 3.9×10 -5 1.45×10 -6 (M) Change +X +X ─X (X in M) Equilibrium X 3.9×10 -5 +X 1.45×10 -6 ─X (M) At equilibrium SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

25. The concentration of un-complexed Ni 2+ = X = 8.93×10 -7 mol/L. The concentration of complexed Ni 2+ = c initial – c equilibrium = 1.45×10 -6 ─ 8.93×10 -7 = 5.6×10 -7 M The percentage of un-complexed Ni 2+ = The percentage of complexed Ni 2+ = Note: The complexed nickel (39%) stays in the environment for a long time because the complexation with HM keeps Ni 2+ ions in solution for a long time. SOLUTION SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

26. Metal complexes with ligands of anthropogenic origin: There are many complexing agents that have anthropogenic sources: a)NH 3 : result from nitrogen-containing organic waste. b)SO 3 2- and SO 4 2- : result from pulp and paper mills. c)PO 4 3- : result from detergents industry. d)CN - : result from gold industry. e)EDTA: result from paper, detergent industry. f)NTA (nitrilotriacetic acid, H 3 T): 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 Cu 2+ ions in solution. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

27. Environmental Chemistry of Colloids and Surfaces SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

28. 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. Colloid 1. SiO 2 2. MnO 2 3. Fe 2 O 3 4. Al 2 O 3 5. Humic material (humic and fulvic acid) 6. Montmorillonite (clay type) 7. Kaolinite (clay type) sediment/water system and soil/water system. some of natural colloids present in natural water. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

29. the size-range distribution of some components of natural water system including colloidal materials. 1.The diameter of colloidal particles falls in the range 0.01 μm to10 μm. 2.Particle size less than 0.45 μm is considered to be soluble in water (true solution) 3.Particle size higher than 0.45 μm is considered to be insoluble(i.e., colloidal solution). SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

30. Why do colloidal materials adsorb different solutes (metals and organics) in solution? SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

31. 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

32. 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. ColloidSpecific surface area (m 2 /g) Montmorillonite5-20 Kaolinite700-800 Humic and fulvic aid700-1000 The specific surface area of some colloids. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

33. 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 m 2 /g. Particle diameter can be calculated from the following relation: Specific surface area (m 2 /g) = 12.5 m 2 /g = 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

34. 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 pH o of the colloid. pH o (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 > pH o then the colloid surface charge is negative and if pH < pH o then the net surface charge is positive and if pH = pH o then the net surface charge is zero (neutral surface). ColloidpH 0 SiO 2 2.0 MnO 2 4.0 Fe 2 O 3 (Hydrated)7.0 Fe 2 O 3 (Goethite)7.5 ColloidpH 0 Fe 2 O 3 (Haematite)8.5 Al 2 O 3 (Hydrated)7.0 Humic material4-5 pH o values of various natural colloids SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

35. 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: 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: Ca 2+ > Mg 2+ > K + > Na + SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

36. 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). SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

37. 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 = pH o of the colloid. The adsorption of organic acids (like fatty acid) by colloidal iron oxides surface is a known example for specific adsorption: Metal cations are also able to form specific bonds with iron-oxide: SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

38. 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. Electrical double layer. 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

39. 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

40. 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: Where; C s : concentration of adsorbate molecules on colloid surface (mol/g) C aq : concentration of adsorbate molecules remaining is solution (mol/L) b: binding constant (L/mol) C sm : maximum quantity of adsorbate adsorbed on the colloid surface (mol/g) SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

41. The linear form of Langmuir equation is: (2) A linear plot between (1/C s versus 1/C aq ) will generate a straight line with (1/bC sm ) as a slope and (1/C sm ) 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 in water C aq (mol/L) in solid (sediment) C s (mol/g) Sample 1: 4.0 × 10 -7 2.0 × 10 -7 Sample 2: 1.3 × 10 -7 1.0 × 10 -7 SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

42. Solution: Applying Langmuir equation to sample 1: Applying Langmuir equation to sample 2: Solving the above linear equations, b and C sm can be obtained: b = 2.7×10 6 L/mol C sm = 3.8×10 -7 mol P/g sediment (or 12.8 μg P/g sediment ). The last number indicate that 12.8 μg of P is adsorbed on one gram sediment. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

43. The Freundlich Equation: Adsorption on colloids or soil can be presented by Freundlich empirical equation: Where: C s : concentration of adsorbate molecules on colloid (mol/g). C aq : concentration of adsorbate molecules remaining is solution (mol/L). K F : Freundlich constant (L/g). n: dimensionless number (n is usually less than 1.0). The linear from of Freundlich equation is: LogC s = LogK F + nLogC aq A linear plot between LogC s versus LogC aq will generate a straight line with (n) as a slope and (LogK F ) as an intercept. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

44. Example For a frost soil, it has been reported that K F = 0.0324 L/g, n = 0.82 for Cd metal adsorption. Calculate C s if C aq = 4 μg/L. Solution: C s = (0.0324 L/g)(4 μg/L) 0.82 C s = 0.1 μg Cd/g 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

45. Partitioning of small organic solutes between water and soil (or colloid surface): 1. The distribution Coefficient K d : At very dilute concentration, n in Freundlich equation can be approximated to unity, therefore, Freundlich equation becomes: Usual units of K d is (L/kg or mL/g) C s : concentration of adsorbate molecules on colloid or soil (mol/kg soil ) C aq : concentration of adsorbate molecules remaining is water (mol/L) Notes If K d >> 1, then the pollutant is mainly adsorbed on the soil If K d << 1, then the pollutant did not adsorb on the soil and remain in water. The value of K d 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 K d for pollutants. In fact, K D value can be correlated to a number of distribution coefficients; K OM, K OC, and K OW. These coefficients can predict the movement of organic pollutants (like pesticides) and inorganic pollutants (like metals) within soil. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

46. 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-NH 2, 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. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

47. 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: C s = f OM C OM + f MM C MM Where f OM and f MM are the fractions of organic matter and mineral matter in the soil. C OM : Concentration of solute, or adsorbate, or pollutant in the organic matter component of the soil or sediment (mol/g OM or mol/Kg OM ). C MM : Concentration of solute, or adsorbate, or pollutant in the mineral matter component of the soil or sediment (mol/g MM or mol/Kg MM ) Assuming that C MM is so small, then: C s = f OM C OM SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

48. 3. Organic Matter-Water Partition Coefficient (K OM ) and Organic Carbon-Water Partition Coefficient (K OC ): The distribution of pollutants in water-soil systems can be described by calculating K OM and K OC : Organic Matter-Water Partition Coefficient: Where; C aq : Concentration of solute, or adsorbate, or pollutant in water (mol/L). C OM :Concentration of solute, or adsorbate, or pollutant in the organic matter component of the soil or sediment (mol/g O.M or mol/Kg O.M ). C OC :Concentration of solute, or adsorbate, or pollutant in the organic carbon component of the soil or sediment (mol/g O.C or mol/Kg O.C ). Note that if K OM and K OC values are greater that 1, then the pollutant is mainly immobilized (adsorbed) on the surface. If the values of K OM and K OC are less that 1, then the pollutant is present in solution and has low affinity for soil. Organic Matter-Water Partition Coefficient: SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

49. 4. The relation between K d, K OM, and K OC coefficients: SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

50. Octanol-Water Partition Coefficient (K OW ): The experimental determination of K d, K OM, and K OC 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 K d, K OM, and K OC 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: K OW value can be used as a guide to predict the movement of pollutants in soils. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

51. 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: Octanol-Water Partition Coefficient (K OW ): Where: K OW is the n-octanol/water distribution coefficient (L water /L octanol ) C octanol : Concentration of solute, or adsorbate, or pollutant in the octanol layer (mol/L octanol ) C aq : Concentration of solute, or adsorbate, or pollutant in aqueous solution (water) (mol/L water ) Note if K OW < 1, then the solute remains in the aqueous phase and has a low molar mass and also has high oxygen content. If K OW > 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 K OW do not adsorb on soil and remain in solution. Solutes of higher values of K OW are highly adsorbed on the soil. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

52. Solutes of higher molar masses have higher K OW SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

53. The relation between K OW and K OM : In fact there is a high correlation between K OW and K OM as shown below: SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

54. 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 SiO 4 2– 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 these minerals has its own CEC capacity. CEC values of some natural clay minerals and some other natural colloids. MaterialMolecular formulaCEC cm(+)/kg Kaolinite * Al 2 Si 2 O 5 (OH) 4 8 Halloysite * Al 2 Si 2 O 5 (OH) 4.2H 2 O8 Smectite * (Na,Ca) 0,3 (Al,Mg) 2 Si 4 O 10 (OH) 2 nH 2 O100 Vermiculite * (Mg,Fe,Al) 3 (Al,Si) 4 O 10 (OH) 10.4H 2 O125 Chlorite * (Fe, Mg, Al)6(Si, Al)4O10(OH)825 HematiteFe 2 O 3 ≈ 4 FeldsparNaAlSi 3 O 8 2 QuartzSiO 2 2 Organic matter See Figure (1) chapter 12200 SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

55. 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: 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: Note that K + is strongly adsorbed compare to Na + on the clay surface. SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

56. Example: 1.0 gram of a clay sample (in Na-form) was mixed with 100.0 mL of 1.0 M of KNO 3 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 this clay sample in cmol(+)/kg clay. 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 + = 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 = 10 2 cmol) = 1.087 × 10 -3 mol (+) = = 0.1087 cmol(+) = 0.1087 cmol(+) Based on that, we can write: 0.1087 cmol(+) → 1.0g ? → 1000.0 g (1.0 kg) CEC ≈ 109 cmol(+)/kg SUMBER: https://eis.hu.edu.jo/ACUploads/10593/Chapter%2012%20new.ppt‎

57. 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