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**Activity and Activity Coefficients**

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**Chemical Equilibrium Electrolyte Effects**

Electrolytes: Substances producing ions in solutions Can electrolytes affect chemical equilibria? (A) “Common Ion Effect” Yes Decreases solubility of BaF2 with NaF F- is the “common ion”

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**The common ion effect is used to decrease the solubility.**

Sulfate concentration is the amount in equilibrium and is equal to the BaSO4 solubility. In absence of excess barium ion, solubility is 10-5 M. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig Predicted effect of excess barium ion on solubility of BaSO4.

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**“inert electrolyte effect”or “diverse ion effect”**

(B) No common ion: “inert electrolyte effect”or “diverse ion effect” Add Na2SO4 to saturated solution of AgCl Increases solubility of AgCl Why??? shielding of dissociated ion species

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**Predicted effect of increased ionic strength on solubility of **

Ksp = Ksp0/fAg+fSO42- Solubility increases with increasing ionic strength as activity coefficients decrease. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Predicted effect of increased ionic strength on solubility of BaSO4. Solubility at zero ionic strength is 1.0 x 10-5 M.

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**Activity and Activity Coefficients**

Activity of an ion, ai = Ciƒi Ci = concentration of the ion ƒi = activity coefficient Ci < 10-4M )= 1 Ionic Strength, m = ½SCiZi2 Zi = charge on each individual ion.

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**Activity and Activity Coefficients**

Calculation of Activity Coefficients Debye-Huckel Equation: -log ƒi = 0.51Zi2(m)½ / (1+0.33ai (m)½) ai = ion size parameter in angstrom (Å) 1 Å = 100 picometers (pm, meters) Limitations: singly charged ions = 3 Å -log ƒi = 0.51Zi2(m)½ / (1+ (m)½)

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**Chemical Equilibria Electrolyte Effects**

Diverse ion (Inert) electrolyte effect For m < 0.1 M, electrolyte effect depends on m only, NOT on the type of electrolyte Solute activities: ax = activity of solute X ax = [X]x x = activity coefficient for X As m 0, x 1, ax [X]

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**Chemical Equilibria Electrolyte Effects**

Diverse Ion (Inert Electrolyte) Effect: Add Na2SO4 to saturated solution of AgCl Ksp = aAg+ . aCl- = 1.75 x 10-10 At high concentration of diverse (inert) electrolyte: higher ionic strength, m aAg+ < [Ag+] ; aCl- < [Cl-] aAg+ . aCl- < [Ag+] [Cl-] Ksp < [Ag+] [Cl-] ; Ksp < [Ag+] = solubility Solubility = [Ag+] > Ksp

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**Diverse Ion Effect on Solubility:**

Presence of diverse ions will increase the solubility of precipitates due to shielding of dissociated ion species. KSP and Activity Coefficients AgCl(s)(AgCl)(aq) Ag+ + Cl- Thermodynamic solubility product KSP KSP = aAg+ . aCl- = [Ag+]ƒAg+. [Cl-]ƒCl- Ḱ SP = [Ag+]. [Cl-] KSP = Ḱ SP ƒAg+. ƒCl- Ḱ SP = KSP/(ƒAg+. ƒCl)

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**Chemical Equilibria Electrolyte Effects**

“Diverse ion (Inert) electrolyte effect” Is dependent on parameter called “ionic strength (m)” m = (1/2) {[A]ZA2 + [B]ZB2 + … + [Y]Zy2} 0.1 M Na2SO4 ; [Na+] = 0.2M [SO4] = 0.1M m = (1/2) {[A]ZA2 + [B]ZB2} m = (1/2) {[0.2](1+)2 + [0.1](2-)2} = 0.3M

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**Chemical Equilibria Electrolyte Effects**

Solute activities: When m is not zero, ax = [X]x Equilibrium effects: mM + xX zZ K =(az)z/(am)m(ax)x K =([Z]Z )z/([M]M )m([X]x )x K ={([Z])z/([M])m([X])x }{Z z/ M m x x} K = K {Z z/ M m x x} Ḱ = K {M m x x / Z z}

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The Diverse Ion Effect The Thermodynamic Equilibrium Constant and Activity Coefficients thermodynamic equilibrium constant, K case extrapolated to infinite dilution At infinite dilution, activity coefficient, ƒ = 1 Dissociation AB A+ + B- K = aA aB/aAB = [A+] ƒA . [B-] ƒB / [AB] ƒAB K = K (ƒA . ƒB / ƒAB) Ḱ = K (ƒAB / ƒA . ƒB )

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**Chemical Equilibria Electrolyte Effects**

Calculation of Activity Coefficients -log ƒx = 0.51Zi2(m)½ / (1+0.33ai (m)½) Where ax = effective diameter of hydrated ion, X (in angstrom units, 10-8cm), Å Ion H3O+ Li+ F- Ca2+ Al3+ Sn4+ ax,, Å 9 6 3.5 11 M 0.86 0.84 0.81 0.48 0.24 0.10

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**Chemical Equilibrium Electrolyte Effects**

Equilibrium calculations using activities: Solubility of PbI2 in 0.1M KNO3 m = 0.1 = {0.1(1+) (1-)2}/2 (ignore Pb2+,I-) ƒPb = ƒI = 0.76 Ksp = (aPb)1(aI)2 = ([Pb2+]Pb )1([I-]I )2 Ksp = ([Pb2+] [I-]2)(Pb I2 ) = Ḱ sp (Pb I2 ) Ḱ sp = Ksp / (Pb I ) Ksp = 7.1 x 10-9 /((0.35)(0.76)2) = 3.5 x 10-8 (s)(2s)2 = Ksp s = (Ksp/4)1/3 s =2.1 x 10-3 M Note: If s = (Kspo/4)1/3 then s =1.2 x 10-3M Solubility calculation difference approx. –43%

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