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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 BaF 2 with NaF F - is the “common ion”

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Fig Predicted effect of excess barium ion on solubility of BaSO 4. The common ion effect is used to decrease the solubility. Sulfate concentration is the amount in equilibrium and is equal to the BaSO 4 solubility. In absence of excess barium ion, solubility is M. The common ion effect is used to decrease the solubility. Sulfate concentration is the amount in equilibrium and is equal to the BaSO 4 solubility. In absence of excess barium ion, solubility is M. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

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(B) No common ion: “inert electrolyte effect”or “diverse ion effect” Add Na 2 SO 4 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 BaSO 4. Solubility at zero ionic strength is 1.0 x M. K sp = K sp 0 /f Ag + f SO42- Solubility increases with increasing ionic strength as activity coefficients decrease. K sp = K sp 0 /f Ag + f SO42- Solubility increases with increasing ionic strength as activity coefficients decrease. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

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Activity and Activity Coefficients Activity of an ion, a i = C i ƒ i C i = concentration of the ion ƒ i = activity coefficient C i < M )= 1 Ionic Strength, = ½ C i Z i 2 Z i = charge on each individual ion.

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Activity and Activity Coefficients Calculation of Activity Coefficients Debye-Huckel Equation: -log ƒ i = 0.51Z i 2 ½ i ½ i = ion size parameter in angstrom ( Å ) 1 Å = 100 picometers (pm, meters) Limitations: singly charged ions = 3 Å -log ƒ i = 0.51Z i 2 ½ ½

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

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Chemical Equilibria Electrolyte Effects Diverse Ion (Inert Electrolyte) Effect: Add Na 2 SO 4 to saturated solution of AgCl K sp = a Ag +. a Cl - = 1.75 x At high concentration of diverse (inert) electrolyte: higher ionic strength, a Ag + < [Ag + ] ; a Cl - < [Cl - ] a Ag +. a Cl - < [Ag + ] [Cl - ] K sp < [Ag + ] [Cl - ] ; K sp < [Ag + ] = solubility Solubility = [Ag + ] > K sp

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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. K SP and Activity Coefficients AgCl (s) (AgCl) (aq) Ag + + Cl - Thermodynamic solubility product K SP K SP = a Ag+. a Cl- = [Ag + ]ƒ Ag+. [Cl - ]ƒ Cl- Ḱ SP = [Ag + ]. [Cl - ] K SP = Ḱ SP ƒ Ag+. ƒ Cl- Ḱ SP = K SP /(ƒ Ag+. ƒ Cl )

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Chemical Equilibria Electrolyte Effects “Diverse ion (Inert) electrolyte effect” Is dependent on parameter called “ionic strength ( ” = (1/2) {[A]Z A 2 + [B]Z B 2 + … + [Y]Z y 2 } 0.1 M Na 2 SO 4 ; [Na + ] = 0.2M [SO 4 ] = 0.1M = (1/2) {[A]Z A 2 + [B]Z B 2 } = (1/2) {[0.2](1+) 2 + [0.1](2-) 2 } = 0.3M

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Chemical Equilibria Electrolyte Effects Solute activities: When is not zero, a x = [X] x Equilibrium effects: mM + xX zZ K =(a z ) z /(a m ) m (a x ) 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 = a A a B /a AB = [A + ] ƒ A. [B - ] ƒ B / [AB] ƒ AB K = K (ƒ A. ƒ B / ƒ AB ) Ḱ = K (ƒ AB / ƒ A. ƒ B )

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Chemical Equilibria Electrolyte Effects Calculation of Activity Coefficients -log ƒ x = 0.51Z i 2 ½ i ½ Where x = effective diameter of hydrated ion, X (in angstrom units, cm), Å IonH3O+H3O+ Li + F-F- Ca 2+ Al 3+ Sn 4+ x,, Å ƒ 0.05 M

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Chemical Equilibrium Electrolyte Effects Equilibrium calculations using activities: Solubility of PbI 2 in 0.1M KNO 3 2 (ignore Pb 2+,I - ) ƒ Pb = 0.35 ƒ I = 0.76 K sp = (a Pb ) 1 (a I ) 2 = ([Pb 2+ ] Pb ) 1 ([I - ] I ) 2 K sp = ([Pb 2+ ] [I - ] 2 )( Pb I 2 ) = Ḱ sp ( Pb I 2 ) Ḱ sp = K sp / ( Pb I ) K sp = 7.1 x /((0.35)(0.76) 2 ) = 3.5 x (s)(2s) 2 = K sp s = (Ksp/4) 1/3 s =2.1 x M Note: If s = (K sp o /4) 1/3 thens =1.2 x M Solubility calculation difference approx. –43%

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