ION ADSORPTION AND EXCHANGE

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ION ADSORPTION AND EXCHANGE

• Cation
• Anion
• Organic Molecules that ionize or
• remain neutral in solution

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Cation Adsorption/Exchange

• Major Cations
• Arid Ca, Mg, K, Na, Zn
• Humid Fe, Al, H, Cu, Mn
• Ion Selectivity
• Strength of adsorption influenced by ion charge,
  charge density, and degree of hydration.
• Lyotrophic Series
• LiNagtKNH4gtRbgtCsMggtCagtSrBagtBaAl

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• Properties
• Nonspecific and based on equivalent reversibility
• Operational Description
• Na-X K K-X Na
• but more correctly described as two separate
  reactions
• Na-X K K-X Na
• K-X Na Na-X K

excess
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• Capacity
• total amount of cations held exchangeably per
  unit mass of soil (mmolc/100g)
• determined by difference (exchangeable
  solution) solution exchangeable
• index cation replacement
• results depend on pH and analytical techniques
  used

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• Given 100 g dry soil
• equilibrate with 60 mL DI water
• Measured Concentrations in Solution
• co, Na 0.006 mmolc/mL
• co, Ca 0.001 mmolc/mL
• co, Cl 0.007 mmolc/mL

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• Extract 100g dry soil with 1N NH4NO3
• Extract Analysis
• TNa 3.4 mmolc/100g
• TCa 22.12 mmolc/100g
• TCl 0.28 mmolc/100g

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Cation Exchange Capacity Calculations

• Estimated By Simple Difference Assumes that the
  anions are expelled from the particle surface and
  are present in solution with an equivalent amount
  of cations as co-ionic complexes.
• ?T ?Tcat - ?Tan
• ?T (3.4 22.12) 0.28 (mmolc/100g)
• 25.52-0.28 25.24 mmolc/100g

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• Summation of Exchangeable Cations Determines the
  amount of individual cations on the exchange by
  difference.
• ?Na TNa w(co,Na)
• w mL/100g co mmolc/mL
• ?Na 3.4 60(0.006) 3.04 mmolc/100g
• ?Ca 22.12 60(0.001) 22.06 mmolc/100g
• ?T 25.10 mmolc/100g
• ?Cl 0.28 60(0.007) -0.14 mmolc/100g

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• By Solution Fractionation Uses mole fraction to
  determine amount in solution.
• ?Na TNa ((co,Na)/(co,Naco,Ca))(TCl)
• ?Na 3.16 mmolc/100g
• ?Ca 22.08 mmolc/100g
• ?T 25.24 mmolc/100g
• Distribution Ratio (RD) ?i/wco,i

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• Chemical Equilibria/Mass Action Equation
• cast in the form of a mass action equation
• Mg-X(s) CaCl2 Ca-X(s) MgCl2
• if the solid phases are treated as pure solids,
  their chemical activities 1, such that
• K (MgCl2)/(CaCl2)
• as long as any Mg or Ca ions are present on the
  exchange sites, the ratio of magnesium to calcium
  chloride activities in solution must remain
  constant

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Cation Exchange Equations

• Kerr
• 2Na-X Ca Ca-X 2Na
• kK Ca-XNa2/Na-X2Ca
• Vanselow
• kV (Ca-X)(Na)2/(Na-X)2(Ca)
• (Ca-X) Ca-X/Ca-X Na-X
• (Na-X) Mg-X/Ca-X Na-X

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• Vanselow
• kV Ca-X/Ca-X Na-X(Na)2
• Na-X/Ca-X Na-X2(Ca)
• Ca-XCa-X Na-X(Na)2
• Na-X2(Ca)
• (Na)2 kV Na-X2
• (Ca) Ca-XCa-X Na-X

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• Gapon (empirical US Salinity Lab)
• Na-X 1/2Ca Ca1/2-X Na
• kG Ca1/2-X(Na)
• Na-X (Ca)1/2
• (Na) kG Na-X (mmol/mL)
• (Ca)1/2 Ca1/2-X

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Field Applications

• The soil is initially at equilibrium.
• Initial status includes a known amount of
  exchangeable monovalent (?) and divalent (?)
  cations, an initial water content (w), and
  initial concentrations of monovalent (c) and
  divalent (c) cations in solution.
• A change is introduced that influences the
  exchange equilibrium.

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• Given
• 100g dry soil
• w 60 mL/100g
• ? 3.0 mmolc/100g
• ? 22.0 mmolc/100g
• c 0.006 mmolc/mL
• c 0.001 mmolc/mL

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• Given the reaction
• Ca1/2-X Na Na-X 1/2Ca
• Initial Equilibrium is characterized by
• ?/?kG(wc/w)/(wc/2w)1/2 (mmolc/mL)
• Summarized in terms of additions of ions and
  water to the solution phase.

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• 1st determine kG for the existing system
• 3.0/22.0 kG (0.006)/(0.06/120)1/2
• (mmolc/100g) (mL/100g)(mmolc/mL)
• kG 0.50
• Introduce a change by adding water (r),
  monovalents (p), and divalents (q), where
• r 140 mL, p 9.6 mmolc, q 0.5 mmolc

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• Left Hand Side of Equation
• (? x)/ (?- x)
• Right Hand Side of Equation
• kG (wc p - x)/(w r)
• ((wc q x)/2(w r))1/2
• Set the LHS RHS, solve for x, and compute the
  new amounts on the exchange and in solution for
  each cation.

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• (? x) kG (wc p - x)/(w r)
• (?- x) ((wc q x)/2(w r))1/2
• NOT SO EASY IS IT??

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Graphical Solution

• 1st assume several arbitrary values for x
• x 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0
• (mmolc)
• Plug in and plot LHS v. RHS where x is along the
  x-axis.
• The intercept between LHS and RHS is the true
  value of x2.0 mmolc
• Solve for ?5.0 and ?20.0 mmolc/100g, and
  c0.04 and c0.0064 mmolc/mL

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HOMEWORK QUESTIONS

• In solving this relationship we used
  concentrations instead of ionic activities does
  that work or did we blow it??
• What if the value of x turns out to be negative
  did we make a mistake??
• What would be the effect of simple dilution (no
  added cations or anions) on the exchange system??
  
• We now have 20 exchangeable Na. Given a soil
  bulk density of 1.3g/cm3, how would you reclaim a
  1 ha area to a depth of 15cm back to 10
  exchangeable Na?? Show all work.

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Special Cases

• Use of the Distribution Ratio (RD)
• If the distribution ratio is very large, i.e.,
  ?i/wci 100, the solution concentration is very
  low in comparison to the exchange capacity (i.e.,
  non-saline with high CEC).
• The maximum relative change in ?i is limited by
  1/RD
• There is nothing more to absorb and the LHS is
  unchanged, so then must be the RHS.
• Complex-dominated System

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• Trace Contaminant Addition
• Actual exchange complex is hardly influenced, but
  what we want to know is what fraction of the
  contaminant remains in solution?
• The exchange equilibrium follows a Kerr-type
  relationship with the dominant ion of the same
  valence.
• ?tc kK (wctc)
• ?Na (wcNa)

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• If we define the initial trace contaminant at
  cotc, the final concentration after introducing
  the spill is then equal to
• ctc (cotc)/(RD 1) (cotc)/(?tc/wctc 1)
• If excessive amounts are added, the solution
  phase becomes dominant, and the system eventually
  adjusts to the composition of the leaching
  solution.
• Solution-dominated System

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Anion Adsorption and Molecular Retention

• Major Anions
• PO43-, SO42-, NO3-, Cl-
• Strength of Adsorption
• PO43- gtgtgt SO42- gtgt NO3- Cl-

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• Theoretical dealings with equilibrium ion
  exchange involves two critical assumptions 1)
  the selectivity constant (k) is truly a constant
  and, 2) the exchange process is reversible.
• The significance of these assumptions becomes
  more apparent when dealing with anion adsorption
  and molecular retention.

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• NonSpecific Adsorption an anion in the vicinity
  of a charged surface may be attracted to
  positively charged sites on the colloidal
  surface, or be repelled by the typical negative
  charge on soil colloids.

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NonSpecific Ion Reactions

• OH-1/2 OH-1/2
• HCl-
• OH-1/2 OH21/2 .Cl-
• OH-1/2 OH-1/2
• NO3 Cl-
• OH21/2 .Cl- OH21/2 ..NO3-

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• Negative Adsorption (Anion Exclusion) When a
  dilute solution of KCl is added to dry smectite,
  the equilibrium Cl concentration is greater than
  the Cl concentration in the original solution.

K
K
Cl-
Dry Soil
HOHKCl
Dry Soil
Wet Soil
K
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• Specific Anion Adsorption solids have the
  ability to specifically interact with various
  anions to the extent that more is adsorbed than
  that predicted from the quantity of negative
  charge required to neutralize surface associated
  positive charge. The theory is that of ligand
  exchange or anion penetration into the actual
  silicate structure itself.

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Specific Anion ReactionsLigand Exchange

• OH21/2 F-1/2
• ..Cl- NaF NaCl
• OH21/2 OH21/2
• OH21/2 PO4H-1.5
• Na NaH2PO4 ...2Na
• OH21/2 OH-1/2

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Specific Anion ReactionsIon Penetration

• Fe-OH Fe-OH Fe-O O
• H2PO4- P
• Fe-OH Fe-H2PO4 Fe-O OH
• OH- HOH

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• Molecular Retention Organic molecules in the
  soil solution can become charged and adsorbed as
  cations or anions, or they may remain nonionic
  and adsorb as a result of polarity within the
  molecule itself.
• Protonation
• B H BH
• BH MSoil BHSoil M

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• Dissociation of weak acids
• HA H A-
• H-Bonding and van Der Waals
• Organic Partitioning

anion retention
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Studied Using Adsorption Isotherms

• Need to characterize the relationship between the
  adsorbed phase (Ca µg/g) and the solution phase
  (Cl µg/mL).
• Cannot utilize equilibrium exchange theory.
• Linear Ca kCl
• Langmuir Ca kbCl/(1kCl)
• Freundlich Ca kCl1/n

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Adsorption Isotherm Development

• Constructed by adding a know amount of solute to
  a known amount of adsorbent.
• Allowing the mix to equilibrate generally by
  shaking for some predetermined period of time.
• Determining the relative concentrations in the
  adsorbed and solution phase.
• Plotting the relationship.

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• Given 5 erlynmeyer flasks, 500 mL each
• Five 25g portions of air dry soil
• w 0.02
• Five 100 mL solute solutions
• 10, 50, 100, 200, and 500 mg/L

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• Amount Applied VolumeConcentration
• 1.0, 5.0, 10.0, 20.0, and 50.0 mg respectively
• Measured Equilbrium Concentrations
• 4.98, 24.9, 49.8, 99.5, and 248.8 mg/L
• Note These are the data points along the x-axis
  (Cl)
• Total Amount of Solution
• 25g soil/(10.02)24.5g dry soil
• 0.5 mL water 100 mL solution 100.5 mL

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• Amount in Solution
• 0.5, 2.5, 5.0, 10.0, and 25.0 mg respectively
• Amount Adsorbed Out of Solution
• 0.5, 2.5, 5.0, 10.0 and 25.0 mg
• Amount Adsorbed on Soil
• 0.5 mg/24.5 g x mg/g convert x µg/g
• 20.41, 102.04, 204.08, 408.16, 1020.40 µg/g
• Note These are the data points along the y-axis.

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THE END