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Ion Exchange Isotherms Models Thermodynamic Exchange Constant Only ions adsorbed as outer-sphere complexes or in the diffuse ion swarm are exchangeable.

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Presentation on theme: "Ion Exchange Isotherms Models Thermodynamic Exchange Constant Only ions adsorbed as outer-sphere complexes or in the diffuse ion swarm are exchangeable."— Presentation transcript:

1 Ion Exchange Isotherms Models Thermodynamic Exchange Constant Only ions adsorbed as outer-sphere complexes or in the diffuse ion swarm are exchangeable Exchange capacities can be determined either at the native pH of soil or at a buffered pH (effective and total exchange capacities, respectively)

2 Exchange Isotherms Typically developed for binary systems Plot charge fraction adsorbed against charge fraction in solution For the surface phase, X i = Z i q i / Q = Z i q i / (z 1 q 1 + z 2 q 2 ) where Z is valance (absolute value if an anion), q is surface excess and Q is exchange capacity For the solution phase, E i = Z i C i / C T where C is solution concentration of charge

3 Homovalent exchange Ca 2+ - Mg 2+ X Ca = 2q Ca / QX Mg = 2q Mg / QX Ca + X Mg = 1 E Ca = 2C Ca / C T E Mg = 2C Mg / C T E Ca + E Mg = 1 If the adsorbent had no preference for either species, it make sense that the isotherm should conform to X Ca = E Ca

4 This can be shown if start with an expression for an exchange equilibrium constant Mg 2+ (ads) + Ca 2+ (aq) = Ca 2+ (ads) + Mg 2+ (aq) K = X Ca (Mg) / X Mg (Ca) which assumes X i accurately models the adsorbed phase activity For non-preference, K = 1 Let’s do some substituting in the above equilibrium expression (Ca) = γ Ca C Ca = E Ca γ Ca C T (Mg) = E Mg γ Mg C T = (1 – E Ca ) γ Mg C T X Mg = (1 – X Ca )

5 Therefore for K = 1, 1 = [X Ca (1 – E Ca ) γ Mg C T ] / [(1 – X Ca )(E Ca γ Ca C T )] [(1 – X Ca )(E Ca γ Ca C T )] = X Ca [(1 – E Ca ) γ Mg C T ] (1 – X Ca ) / X Ca = (1 – E Ca ) / E Ca since γ Ca = γ Mg and X Ca = E Ca Example of nearly non-preference Ca – Mg exchange on 2:1 mineral. Dominant surface was Si tetrahedral sheet with diffuse charge. Use of ClO 4 - avoided solution complexes.

6 The same result is obtained if instead of modeling activity of adsorbed species by X, it is modeled by mole fraction, N, on the surface where N 1 = q 1 / (q 1 +q 2 ) and N 2 = q 2 / (q 1 + q 2 ) This is obviously true for homovalent exchange since in this case N i = X i, however, for heterovalent exchange, i.e., Ca – Na, the expressions are different.

7 Deviation from non-preference Ca – Mg exchange in mixed mineralogy system

8 For the exchange reaction, 2Na + (ads) + Ca 2+ (aq) = Ca 2+ (ads) + 2Na + (aq) for which K = N Ca (Na) 2 / N Na 2 (Ca) i.e., surface phase activities modeled as mole fractions Derive X Ca = F(E Ca ) in the case of non-preference exchange, i.e., K = 1 This is less straightforward but start with the substitutions N Ca = q Ca / (q Ca + q Na )X Ca = 2q Ca / QX Na = q Na / Q N Ca = X Ca / (X Ca + 2X Na ) = X Ca / (2 - X Ca ) since X Ca + X Na = 1 N Na = q Na / (q Ca + q Na ) = 2(1 - X Ca ) / (2 - X Ca ) C Ca = E Ca C T / 2 C Na = E Na C T = (1 – E Ca )C T since E Ca + E Na = 1 where C T = 2C Ca + C Na

9 Substituting in terms of X Ca and E Ca and including γ i s 1 = [X Ca (1 – E Ca ) 2 C T 2 γ Na 2 ] / {[2(1 – X Ca ) 2 / (2 – X Ca )] (E Ca C T γ Ca )} which rearranges to X Ca 2 - 2X Ca + 2 / (1 - E Ca ) 2 C T 2γ Na 2 / E Ca γ Ca = 0 from which X Ca = 1 - [β / (1 + β)] 1/2 where β = (1 - E Ca ) 2 C T γ Na 2 / 2E Ca γ Ca and ranges from  to 0 E Ca = 0, X Ca = 0 E Ca = 1, X Ca = 1 E Ca = y, X Ca > y Can show this by substitution or dX Ca / dE Ca at E Ca = 0 and E Ca = 1

10 Deviation from non-preference in heterovalent exchange occurs even with 2:1 minerals dominated by Si tetrahedral surface. Non-preference isotherm not shown but would lie below (Ca – Na) or above (Na – Mg) data.

11 5. Given the below exchange data for solution, mNa and mMg, and adsorbed, qNa and qMg, phases, graph the exchange isotherm, examine applicability of the non-preference isotherm and compare it with a fitted isotherm. mNa mMg qNa qMg ------ mol / kg ------ - mol / kg - 0.04950 0.00117 0.53 0.28 E Na = m Na / (m Na + 2m Mg ) = m Na / C T 0.04740 0.00234 0.30 0.45 0.04400 0.00700 0.22 0.70 E Mg = 2m Mg / C T 0.03830 0.00940 0.23 0.86 0.03400 0.01240 0.10 0.74 X Na = q Na / (q Na + 2q Mg ) = q Na / Q 0.02910 0.01490 0.08 0.74 0.02370 0.01740 0.06 0.78 X Mg = 2q Mg / Q 0.01850 0.01970 0.06 0.95

12 ENa EMg XNa XMg γNa γMg XMg-NP SQErr1 SQErr2 R2 0.9549 0.0451 0.6543 0.3457 0.8206 0.4534 0.2254 0.1788 0.0000 0.9101 0.0899 0.4000 0.6000 0.8206 0.4534 0.3625 0.0284 0.0110 0.7586 0.2414 0.2391 0.7609 0.8206 0.4534 0.6122 0.0001 0.0021 0.6708 0.3292 0.2110 0.7890 0.8206 0.4534 0.6965 0.0004 0.0001 0.5782 0.4218 0.1190 0.8810 0.8206 0.4534 0.7642 0.0126 0.0025 0.4941 0.5059 0.0976 0.9024 0.8206 0.4534 0.8140 0.0179 0.0012 0.4051 0.5949 0.0714 0.9286 0.8206 0.4534 0.8583 0.0256 0.0009 0.3195 0.6805 0.0594 0.9406 0.8206 0.4534 0.8950 0.0296 0.0002 0.7685 0.2935 0.0180 0.94

13 K Na Mg XMgAltSQErr2R2R2 1.0390.42150.0057 1.8720.56790.0010 1.6930.76330.0000 1.2430.81860.0009 2.4080.86090.0004 2.1960.89100.0001 2.3370.91740.0001 1.8210.93900.0000 1.4910.00830.97 The form of conditional exchange constant used mole fraction to model surface phase activities. Vanselow, K V.

14 Exchange Models These equilibrium expressions are referred to as selectivity coefficients. Largely differ based on how surface phase activities are approximated. Either as a function of equivalent fraction or mole fraction on the adsorbent, (A Z+ ads ) = X A F(Z) or (A Z+ ads ) = N A exp(F(N A, N B )), where B is the other cation in the binary exchange. The objective in modeling surface phase activities is to best describe the exchange equilibria across the full range of surface phase compositions using a single value, a constant. This value would, therefore, approximate the thermodynamic exchange constant.

15 (A Z+ ads ) = X A F(Z) Gaines-Thomas Model surface phase activities as equivalent fractions directly, for example, K = X Ca (Na) 2 / X Na 2 (Ca) where X Ca = 2q Ca / (2q Ca + q Na ) = 2q Ca / Q X Na = q Na / (2q Ca + q Na ) = q Na / Q and X Ca + X Na = 1 In this case, (A Z+ ads ) = X A F(Z) = X A, i.e., F(Z) = 1 Notice that this form is very close to the exchange isotherm data and if isotherm data were used to calculate k at each point, k = K Gaines-Thomas (2 / C T ) (γ Ca / γ Na 2 ), for this heterovalent exange and k = K Gaines-Thomas for homovalent exchange.

16 XMgAltSQErr1SQErr2R2R2 K Na Mg 0.38430.17880.0015 2.511 0.54440.02840.0031 5.347 0.74030.00010.0004 5.465 0.80270.00040.0002 4.107 0.84660.01260.0012 8.606 0.87980.01790.0005 8.003 0.90920.02560.0004 8.725 0.93330.02960.0001 6.877 0.29350.00730.974.490 The form of conditional exchange constant used charge fraction to model surface phase activities. Gaines-Thomas, K GT.

17 Gapon The exchange reaction may be written ½ Ca 2+ (aq) + Na + (ads) = ½ Ca 2+ (ads) + Na + (aq) for which K (X Ca ) 1/2 = X Ca (Na + ) / X Na (Ca 2+ ) 1/2 = K Gapon If (A Z+ ads ) = X A Z, then (Ca ads ) = X Ca 2 and (Na ads ) = X Na which gives K Gapon = X Ca (Na + ) / X Na (Ca 2+ ) 1/2 Thus, K Gapon = X Ca 1/2 (K Gaines-Thomas ) 1/2

18 (A Z+ ads ) = N A exp(F(N A, N B )) Vanselow Simplest among such models with F(N A, N B ) = 0, thus, surface phase activities are modeled directly as mole fractions 2Na + (ads) + Ca 2+ (aq) = Ca 2+ (ads) + 2Na + (aq) K = N Ca (Na) 2 / N Na 2 (Ca) = 1 where N Ca = q Ca / (q Ca + q Na ) N Na = q Na / (q Ca + q Na )and N Ca + N Na = 1 This form of a conditional exchange constant (selectivity coefficient) can be manipulated in such way as to give a thermodynamic exchange constant based on exchange isotherm data. May furthermore calculate ΔG o, ΔH o and ΔS o for the exchange reaction. See handout.

19 K Na Mg XMglnK V dX Mg 1.0390.34570.0131 1.8720.60000.1594 1.6930.76090.0847 1.2430.78900.0061 2.4080.88100.0808 2.1960.90240.0169 2.3370.92860.0222 1.8210.94060.0072 1.4910.0356 0.4260 SUM 1.5312EXP(SUM)K 1 ln K =  ln K V dX B 0 Extrapolated to X Mg = 1 with K V = 1.821

20 ln K V = -0.278 + 1.097X Mg R 2 = 0.56 1 ln K =  ln K V dX B = 0.270 and K = 1.310 0

21 Since ΔG o = ΔH o – TΔS o = -RT ln K R ln K = - ΔH o / T + ΔS o if exchange experiment done at two temperatures, R ln K T2 – R ln K T1 = -ΔH o / T 1 + ΔH o / T 2 = ΔH o (1 / T 2 – 1 / T 1 ) ΔH o = R ln (K T2 / K T1 ) x T 1 T 2 / (T 1 – T 2 ) ΔS o = R ln K + ΔH o / T


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