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Soil chemistry

  1. 1. 1 Mid Exam SOIL CHEMISTRY SES-701 Q1. Differentiate between ion pair and complex ions and explain the agricultural significance of ion pair formation? Ion pair: Ligands attached at outer solvation sphere of the central cations, weak electrostatic forces. Complex ion: The ligands enter in the inner solvation sphere of the central cations and attach, and it remove water molecules. E.g H2O in Al (H2O))3+6 and Fe (H2O) 63+ In complex ion, for association with central cation, ligands must compete with water molecule in central ion’s solvation and must lose its water molecule. In case of ion pair ligand attach outside inner solvation sphere M + A - Ion Pair + M A - + M A - Complex Ions A ligand is any ion or molecule in the coordination sphere of the central ion. Agricultural significance: Complex ions lower the activity of catons adsorption and effective concentration of free cations, toxicity is prolonged, lowers the leaching of ions. Apparent super saturation of CaCO3o in ground water and drainage water in arid regions. Predominantly ion pair form in arid and semi-arid soils are CO3, HCO3 and sulphate. Improves the knowledge about solid phase which ultimately controls ion concentration in soil solution Q2. Define the following: a) Activity coefficient: The factor by which the value of a concentration of a solute must be multiplied to determine its true thermodynamic activity. Or Activity coefficient is a measure of how much the solution differs from an ideal solution. M.Nadeem Ashraf (Hons) Soil Science UAF
  2. 2. 2 b) Physical adsorption: This process cause a little or no disruption of the electron structure of the adsorbate molecules. Its also called Van der Waals adsorption and non specific process. c) Chemisorption: it results from the interaction of electron shells of molecules forming primary valence bonds between the surface and adsorbate molecules. It is specific adsorption in which cation embeds and becomes part of the system and lose their specific identities. Q3. Write down the Nernst equation and briefly define its components explain it agriculture significance? Ψo = ln Ψo = potential at zero distance, K= Boltzman constant, T= absolute temperature Z= valency of ion, e= electron charge (1.6 x 10-19 C) , a+o, a- = activities of potential determining ion at zero distance Use of Nernst equation: Nernst equation relates the DDL potential with the activity of potential determining ion. Q4. Enlist the factors affecting cation exchange, and elaborate the valence dilution effect along with example? OR Assumptions/principles of exchange reactions?  Reversibility  Stoichiometry  Speed  Mass action  Valence dilution  Complementary cations/anions  Colloid specific effect  CEC of colloids Limitations of exchange reactions:  Simultaneous exchange of cations and anions is never considered but does occur.  Constant CEC or AEC is assumed, this is valid for some but not for all situations. M.Nadeem Ashraf (Hons) Soil Science UAF
  3. 3. 3  Simple stoichiometric reactions  Complete reversibility in exchange reaction Q5. Explain valence dilution effect with numerical example: Valence dilution effect: For exchange between cations of unequal valence “dilution of the equilibrium solution favours retention of the more highly charged cations at the cost of monovalent” Cax2+ 2NH4+ (NH4+) 2X + Ca2+ K= [( ) [ ][ ][ ] _____(i) ] Here brackets shows concentrations ( moles/litre) , rearranging the equation as under: [( ) [ ] [ ] ] [ ] _____(ii) Because of the squared term on the right side of the equation(ii), the ratio of NH 4+ to Ca2+ in the colloids double layer changes with total, as well as relative salt concentration of bulk solution. This dependence of cation exchange on cation valence is called the valence dilution effect. Numerical example: consider a solution having [NH4+]=[Ca2+]=1 mmol/litre, the ratio [NH4+]/[Ca2+] in this case equals to [1]2/[1] or 1mmol/litre. Upon 10 time dilution, the ratio becomes [0.1]2/[0.1] or 0.1 mmol/litre. Hence, the ratio of [NH4+]/[Ca2+] on the colloid will decrease during the dilutions. Q6. Differentiate between kaolinite and vermiculite. Kaolinite: One octahedral sheet is attached with one tetrahedral sheet to form 1:1 layer silicate e.g. Kaolin Group (Kaolinite, Halloysite)  Layers are attached with each other by H-bonds  Kaolinite is a non-expanding mineral,  Unable to absorb water into the interlayer position  interlayer spacing fixed at 0.72 nm  95% variable charge  Surface area is 10-20 m2 g-1 M.Nadeem Ashraf (Hons) Soil Science UAF
  4. 4. 4 Vermiculite: One octahedral sheet is attached with two tetrahedral sheet to form 2:1 layer silicate.      Vermiculite is an expanding mineral Swelling, plasticity and shrinkage properties 95% charge is through isomorphic substitution. Zero variable charge CEC is 100-150 cmol (+) kg-1. Specific surface area is 600-800 m2g-1. Montmorillonite: Expanding mineral (2:1 type) CEC is 80-120 cmolc kg-1 Specific surface area is 600-800 m2g-1 Swelling, plasticity and shrinkage properties well expressed but lees than vermiculite  Particle size is 0.01-1.0 µ ( spacing 12-18Ao)  O-O bond which is so weak that each unit can be separated from others.     Mica:          Inherited from parent material Non-expanding (2:1 type) Maximum layer charge but unavailable K is fixed between two layers through K-O CEC is 5-15 mmolckg-1 Dominant in silt and sand fractions C-spacing is 10Ao Specific surface area is 70-120 m2g-1 Dominant in Pakistan soils Q7. Define the following a) Tetrahedral coordination: It consists of four O2- ligands coordinated around one Si4+ giving an ionic unit SiO44-. b) Octahedral coordination: It consists of six O2- or OH- groups coordinated around a central cation of Al3+ c) Solubility product constant: It refers to ion activities in equilibrium with their solid phases, M.Nadeem Ashraf (Hons) Soil Science UAF
  5. 5. 5 d) STABILITY CONSTANT It refers to the complex ion and ion pair, Formation of complex ion is due to cation-anion attractive forces that overcome attractive forces between cation and H+ for various ligands including water Al(H2O)63+ + F- ======== AlF(H2O)52+ + H2O E) Tri-octahedral coordination: In octahedral coordination, when all the positions of the central cavity are occupied by a divalent cation e.g. Talc F ) Di-octahedral coordination: In octahedral coordination, when 2 out of 3 positions of the central cavity are occupied by a trivalent cation, e.g. pyrophylite Q8. Define Complementary Ion Effect and write down factors affecting CIE? Complementary Ion Effect It is defined as the influence of one adsorbed ion on the release of another from the surface of a colloid. Factors Affecting Complementary Ion Effect 1. Type of colloid: High CEC colloids tend to preferably retain polyvalent cations and vice versa. 2. Valency of ions: High valent ions have preferential adsorption and vice versa. 3. Size of ions: Smaller size ions are preferred for adsorption and vice versa. 4. Temperature: Increasing temperature tends to increase the exchange of ions and vice versa. 5. Dilution: Dilution favors retention of polyvalent ions while concentration tends to reverse it. 6. Hydrated ionic size: Larger size ions (hydrated) tend to be replaced easily and vice versa. 7. Concentration of desorbed ions and their removal from the system: Low concentration in solution and / or removal of ions from solution favors desorption and vice versa. 8. Activity coefficient of replaced and complementary ions M.Nadeem Ashraf (Hons) Soil Science UAF
  6. 6. 6 Q9. Write a short note on lyotropic series. Ease with which different cations replaced and held on exchange surface of colloid monovalent is less tightly than divalent. With a given valence series, the degree of replacibility of an ions decreases as its dehydrated radius increases except H+. Ions of smaller dehydrated radius have a greater density of charge per unit volume; hence they attract water of hydration more strongly. The radius of such hydrated cations is large than dehydrated. Q10. Potential Determining ion Ion that determine potential of surface by entering into specific reaction with one of the constituents of the solid. AgI + I SiO2 + OH - - AgI - - SiO2 (OH) I- and OH- are potential determining ions. Q11. Why pH dependent charge is dominant in kaolinite clay mineral. pH dependent charge develops due to functional groups on surface of soil colloid, so in kaolinite dissociation of Si-OH and Al-OH groups on the crystal edge is the principal source. Kaolinite has positive edge charge in acidic medium and negative edge charge in alkaline medium. Q12. Define the following terms: a) Electrostatic potential energy: work is associated with moving an ion into the force field(DDL) of an interface, the ion must have some potential energy by nature of its position in the DDL. The electrostatic potential of a charge q which is located in an electric field at x position is: EPE= qΨx = ZeΨx b) Thermal energy of an ion: it is the product of absolute temperature and Boltzman constant. T.E= KBT c) Ionic strength: It is measure of intensity of electric field due to the ions in solution of an electrolyte. I= ∑CiZi2 For the natural water, ionic strength can be calculated from EC (dSm-1) as: I= 0.0127 EC r=0.996 (Griffin and Jurinack in soil science 116:26) M.Nadeem Ashraf (Hons) Soil Science UAF
  7. 7. 7 I= 0.0129 EC r=0.997 (Ghafoor et al; 2000) Examples: 1. Calculate ionic strength for 0.1 m solution of CaCl2, Ca and Cl also calculate activity coefficient for Ca and Cl I (CaCl2) = ½ Σ (C1Z12 + C2Z22……CnZn2) I (CaCl2) = ½ Σ (0.1 X 22 + 0.2 X 12) = 0.30 moles/L I (Ca) = ½ Σ (0.1 X 22) I (Cl2) = ½ Σ (0.2 X 12) = 0.20 mole/L = 0.10 moles/L Activity coefficient Debye-Huckel equation -log (γ) = AZ2I1/2 -log (γ)CaCl2 = 0.511 (2) (1) (0.30)1/2 =0.55 -log (γ)Ca =0.511 (2)2(0.20)1/2 =0.89 -log (γ)Cl = -0.511 (1)2 (0.10)1/2 =0.10 2. Calculate ionic strength of 0.01 M Nacl, CaCl2, and Al2 (SO4)3 solutions (do yourself). Q13. Differential between rocks and minerals and write down their uses. Rocks: A natural aggregate of one or more minerals to form an appreciable part of solid portions of earth. i) ii) iii) Igneous rocks: igneous rocks are formed by solidification of molten material called magma, which originates at considerable depth beneath at the earth’s surface. Sedimentary rocks: are composed of particles derived from previously existing rock, then deposited after transportation by streams, waves and wind or ice. Metamorphic rocks: are derived from the pre-existing igneous or sedimentary rocks by alteration due to the high pressure and temperature that accompany mountain-building movements of earth crust. M.Nadeem Ashraf (Hons) Soil Science UAF
  8. 8. 8 Uses of Rocks: • Stone age→ Bronze age → Iron age • Construction of buildings and infrastructure → Dimension stone • Source of essential plant nutrients released upon weathering (over periods of time) • Valuable minerals and rocks → industrial minerals • Rocks from which minerals are extracted for economic purposes are referred to as ores. Q14. Differentiate between primary and secondary minerals? Primary minerals: a mineral that has not been altered chemically from deposition and crystallization from the molten lava. E.g. quartz Secondary minerals: a mineral resulting from the decomposition of primary minerals. E.g. calcite, gypsum. Q15. Write down the general form of Boltzmann equation and define its components. ni = nio exp ( ) ni : is Conc. of ion 'i' at point x as ions cm-3 , T : Absolute temperature nio : Conc. of ion 'i' in the bulk solution as ions cm-3 , e = Electron charge exp = 2.7183 and is constant, ψx : Electric potential at distance x, zi : Valency of ion ‘i’, k=Boltzman (1884) constant = R/NA Q16. Enlist the factors which affect the thickness of DDL and define zeta potential. Factors Affecting DDL Thickness: (i) Electrolyte concentration (ii) Surface charge density (iii) Valence of counter ions (iv) Temperature. M.Nadeem Ashraf (Hons) Soil Science UAF
  9. 9. 9 Effect of the first two factors (i and ii) on the DDL thickness in aqueous solution at room temperature is combined equation: K = 3 × 107z(c)1/2 Where K is reciprocal of DDL thickness, ‘z’ is the valency of counter ion and ‘c’ is electrolyte concentration in the bulk solution as mol L-1. The DDL thickness is thus: 1/K = 3 × 10-7 / z (c)1/2 cm Zeta Potential (Ψz): The relative motion of a charged particle with respect to the bulk solution. Q17. Why there is an increase in salinity of soil solution, decreases in the apparent CEC? The salinity of soil solution increases, the apparent CEC decreases due to blocking effect of cation on the availability of negative charge on clays. The CEC increases on dilution and approaches the amount of isomorphic substitution as an upper limit. Q18. Why anion concentration (e.g. Cl) in soil extract is often higher than the calculated amount? Anion concentration of chloride ion is higher in soil extract due to more solubility of it’s than carbonate and bi-carbonate. Q19. Describe DDL theory and write down DDL assumption. DDL theory The concentration of counter ions is the highest in the immediate vicinity of a surface and decreases at first rapidly and then asymptotically to the intermicellar solution of uniform ionic concentration. The potential at the surface is ψ0 which decreases with distance from the surface. M.Nadeem Ashraf (Hons) Soil Science UAF
  10. 10. 10 Assumptions: (i) (ii) (iii) (iv) Ions interact electro-statistically like charged particles of zero size but in reality ions and their associated water molecules have appreciable physical size. Exchangeable cations exist as point charges. Colloidal surfaces are planner and essentially infinite in extent. Surface charge is distributed uniformly over the entire surface of a colloid. Q20. Describe zeta potential and enlist the methods for its measurement. Zeta potential:  It may be defined as the relative motion of a charged particle with respect to the bulk solution.  Although the ion atmosphere of the DDL is bound to the charged surface by electric forces which prevent its escape into bulk solution, a portion of DDL is within the free solvent and hence not held with great tenacity.  Indeed, a certain portion of the DDL can be made to move along the surface if an appropriate force is applied.  An outside electric force applied to a colloid system can cause a separation of the DDL with a certain amount of solvent and counter ions fixed to the charge surface and the remaining portion of the DDL moving as free solvent (water).  The plane of separation of the DDL into its ‘bound’ and ‘free’ components is called the plane of shear. The electric surface potential which exists in the DDL at the plane of shear is called the Ψz. Determination of Zeta Potential (Ψz)  Electrophoresis  Electro-osmosis:  Streaming Potential M.Nadeem Ashraf (Hons) Soil Science UAF
  11. 11. 11 Significance of Zeta Potential:  Its value can be related to the stability of colloidal dispersions.  It indicates the degree of repulsion between adjacent similarly charged particles, i.e. degree of dispersion.  For molecules and particles that are small enough, a high Ψz will confer stability, i.e. the solution or dispersion will resist aggregation.  When the potential is low, attraction exceeds repulsion and the dispersion will break and flocculate.  Colloids with high Ψz (negative or positive) are electrically stabilized while colloids with low Ψz tend to coagulate.  The Ψz is widely used for quantification of the magnitude of the electrical charge at the DDL.  Ψz is often the only available path for characterization of DDL properties. Q21. Write down the modification proposed by Stern in the DDL model? Stern (1924) introduced two major corrections to the Gouy−Chapman model. (i) Finite dimensions of counter ions in the region adjacent to the charged surface were corrected. (ii) Possibility of the specific adsorption of counter ions by the surface was included. According to Stern, the counter ions can be considered as divided between diffuse layer of point charges and an immobile surface layer of some thickness, which can contain a certain maximum number of counter ions cm-2 of the interface. M.Nadeem Ashraf (Hons) Soil Science UAF
  12. 12. 12 Q22. Write down the assumption of Langmuir equation and enlist other ion exchange equation? Langmuir assumptions: (i) (ii) (iii) The surface is energetically homogeneous; all with the adsorbed molecules at the same level ΔHads is constant for the total surface. The presence of an adsorbate molecule on a influence neighboring surface sites. The adsorbed molecules do not interact adsorbateadsorbate interaction. sites interact of energy, i.e. site does not laterally, no Ion exchange equations: i.Double Layer theory: Gouy-Chapman theory and Stern Layer theory. ii. Donnon Equilibria, iii. Statistical Thermodynamics: Davis Equation iv. Based on Mass Action: Gapon Equation, Vanselow Equation, and Kerr Equation. v. Kinetic Approach: Freundlich and Langmuir Equations M.Nadeem Ashraf (Hons) Soil Science UAF