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Activity and Activity Coefficients
Chemical Equilibrium Electrolyte Effects
Electrolytes:
Substances producing ions in solutions
Can electrolytes affect chemical equilibria?
(A“ )Common Ion Effect” YesDecreases solubility of BaF2 with NaF
F- is the “common ion”
Fig. 10.3. Predicted effect of excess barium ion on solubility of BaSO4.
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.
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)
(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 ofBaSO4. Solubility at zero ionic strength is 1.0 x 10-5 M.
Ksp = Ksp0/fAg+fSO42-
Solubility increases with increasing ionic strength as activity coefficients decrease.
Ksp = Ksp0/fAg+fSO42-
Solubility increases with increasing ionic strength as activity coefficients decrease.
©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)
Activity and Activity Coefficients
Activity of an ion, ai = Ciƒi
Ci = concentration of the ion
ƒi = activity coefficient ) @ Ci < 10-4M (= 1
Ionic Strength, = ½CiZi2
Zi = charge on each individual ion.
Activity and Activity Coefficients
Calculation of Activity Coefficients
Debye-Huckel Equation:
-log ƒi = 0.51Zi2½i ½
i = ion size parameter in angstrom )Å(
1 Å = 100 picometers )pm, 10-10 meters(
Limitations: singly charged ions = 3 Å
-log ƒi = 0.51Zi2½ ½
Chemical Equilibria Electrolyte Effects
Diverse ion )Inert( electrolyte effectFor < 0.1 M, electrolyte effect depends on only, NOT on the type of electrolyte
Solute activities:
ax = activity of solute X
ax = [X]x
x = activity coefficient for X
As 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,
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-
K SP = [Ag+]. [Cl-]
KSP = K SP
ƒAg+. ƒCl-
K SP = KSP/)ƒAg+. ƒCl(
Chemical Equilibria Electrolyte Effects
“Diverse ion )Inert( electrolyte effect”
Is dependent on parameter called “ionic strength )”
= )1/2( {[A]ZA2 + [B]ZB
2 + … + [Y]Zy2}
0.1 M Na2SO4 ; [Na+] = 0.2M [SO4] = 0.1M
= )1/2( {[A]ZA2 + [B]ZB
2}
= )1/2( {[0.2])1+(2 + [0.1])2-(2} = 0.3M
Chemical Equilibria Electrolyte Effects
Solute activities:
When 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}
K = K {M m x
x / Z z}
The Diverse Ion Effect
The Thermodynamic Equilibrium Constant and Activity Coefficientsthermodynamic 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(
K = K )ƒAB / ƒA . ƒB (
Chemical Equilibria Electrolyte Effects
•Calculation of Activity Coefficients
•-log ƒx = 0.51Zi2½i ½
•Where x = effective diameter of hydrated ion, X )in angstrom units, 10-8cm(, Å
• IonH3O+Li+F-Ca2+Al3+Sn4+
x,, Å963.56911
ƒx @ 0.05
M
0.860.840.810.480.240.10
Chemical Equilibrium Electrolyte Effects
•Equilibrium calculations using activities:
Solubility of PbI2 in 0.1M KNO3
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
( = K sp (Pb I2
(
K 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)1/3s =2.1 x 10-3 M
Note: If s = (Kspo/4)1/3 thens =1.2 x 10-3M
Solubility calculation difference approx. –43%
