Activity and Activity Coefficients

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”

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. 10.3. Predicted effect of excess barium ion on solubility of BaSO4.

“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

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.

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.

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, 10-10 meters) Limitations: singly charged ions = 3 Å -log ƒi = 0.51Zi2(m)½ / (1+ (m)½)

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]

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

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)

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

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}

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 )

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 ƒx @ 0.05 M 0.86 0.84 0.81 0.48 0.24 0.10

Chemical Equilibrium Electrolyte Effects Equilibrium calculations using activities: Solubility of PbI2 in 0.1M KNO3 m = 0.1 = {0.1(1+)2 + 0.1(1-)2}/2 (ignore Pb2+,I-) ƒPb = 0.35 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%