1. Activity and Activity Coefficients

2. 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”

3. 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.

4. “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

5. 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.

6. 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.

7. 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)

8. 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]

9. 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

10. 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)

11. 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

12. 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}

13. 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 )

15. 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

16. 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%