CHEMISTRY 59-320
ANALYTICAL CHEMISTRY
Fall - 2010
Chapter 8: Activity and the systematic treatment of equilibrium

Chemical Equilibrium Electrolyte Effects
•
The right hand side figure shows that the equilibrium constant of (8-1) decreases as electrolyte that does not participate the reaction is added (Why?).
• Electrolyte: Substances producing ions in solutions.
• Can electrolytes affect chemical equilibria?
(A) “Common Ion Effect”  Yes
Decreases solubility of BaF
2
F is the “common ion” with NaF
(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 does the solubility increase when salts are added to the solution?
• The formation of ionic atmosphere. The greater the ionic strength of a solution, the higher the charge in the ionic atmosphere.
• The ionic atmosphere attenuates the attraction between ions since each ionplus-atmosphere contains less net charge.
• Increasing ionic strength therefore reduces the attraction between any particular Ag + and any Cl , relative to the case in distilled water.

• Ionic strength, μ, is a measure of the total concentration of ions in solution.
m
= ½ S
C i
Z i
2 where C i is the concentration of the ion, actually a ratio of C
= charge on each individual ion.
i
/1 M, Z i
Example: Find the ionic strength of (a) 0.10 M NaCl; (b) 0.020 M KBr plus 0.01
M Na
2
SO
4
.
Solution:
(a) the soultion contains 0.10 M Na + and 0.10 M Cl m
= ½ [(0.10M/1M)*(+1) 2 + (0.10M/1M)*(-1) 2 ] = ½ [0.10 + 0.10] = 0.10
(b) the solution contains 0.020 M K + , 0.020 M Br , 0.02 M Na + and 0.01 M
SO
4
2.
m
= ½ [(0.020M/1M)*(+1) 2 + (0.020M/1M)*(-1) 2 + (0.020M/1M)*(+1) 2 +
(0.010M/1M)*(-2) 2 = ½ [0.020 + 0.020 + 0.020 + 0.040] = 0.050.
Ions with a larger charge number have greater contribution on the ionic strength .

• To account for the effect of ionic strength, concentrations in the calculation of equilibrium constant show be replaced by activities: Ẳ
C
= [C]* γ
C
• The activity of species j is its concentration multiplied by its activity coefficient, a i
= C i
ƒ i
ƒ i
= activity coefficient.
• At low ionic strength, activity coefficients approach unity.

• Calculation of Activity Coefficients
• Extended Debye-Huckel Equation:
• a i
= ion size parameter in angstrom ( Å )
1 Å = 100 picometers (pm, 10 -10 meters)
• Limitations: singly charged ions = 3 Å
• log ƒ i
= - 0.51Z
i
2
(m) ½
/ (1+ (m) ½ )
• On page 144: “the equation works fairly well for μ≤0.1 M “ (??)

Effect of ionic strength, ion charge, and ion size on the activity coefficient
• The ion size α in eq 8-6 is an empirical parameter that provides good agreement between measured activity coefficients.
• α is the diameter of the hydrated ion .
• As ionic strength increases, the activity coefficient decreases.
• The activity coefficient approaches unity as the ionic strength approaches 0.
• As the magnitude of the charge of the ion increases, the departure of its activity coefficient from unity increases.
• The smaller the ion size ( α), the more important activity effects become.

Activity coefficients for differently charged ions with a constant ionic size

• Case 1: neutral molecules in solution phase, the activity coefficient is unity and thus the activity is numerically the same as their concentration.
• Case 2: Gases. The fugacity (i.e. activity) is calculated as the product of pressure and the fugacity coefficient (i.e. activity coefficient).
When the pressure is below 1 bar, the fugacity coefficient is close to unity.

• K sp o
= a
Ag +
. a
Cl =
1.75 x 10 -10
• Adding Na
2
SO
4 to saturated solution of AgCl, at high concentration of diverse (inert) electrolyte: higher ionic strength, m
• a
Ag +
< [Ag + ] ; a
Cl -
< [Cl ]
• a
Ag +
. a
Cl -
< [Ag + ] [Cl ]
• K sp o < [Ag + ] [Cl ] ;
 K sp o < [Ag + ] = solubility
• Solubility = [Ag + ] >

K sp o

• Solubility of PbI
2 in 0.1M KNO
3
• m = 0.1 = {0.1(1 + ) 2 + 0.1(1 ) 2 }/
2 (ignore Pb 2+ ,I )
•
•
•
• ƒ
Pb
= 0.35
ƒ
I
= 0.76
K
K
K sp o sp o sp
= (a
Pb
) 1 (a
I
= (
= K
[Pb sp o
2+ ] [I
) 2 = ( [Pb 2+ ]

Pb
) 1 ( [I ]

I
) 2
]
/ (

Pb
2 )(

Pb

I
)

I
2 ) = K sp
(

Pb

I
2 )
• K sp
= 7.1 x 10 -9 /(( 0.35)(0.76) 2 ) = 3.5 x 10 -8
• (s)(2s) 2 = K sp s = (Ksp/4)
• Note: If s = (K sp o /4) 1/3 then
1/3 s =2.1 x 10 s =1.2 x 10
-3
-3
M
M
• The solubility is increased by approx. 43%

Addition example: Calculate the pH of water containing 0.010 M KCl at 25 o C.
(see in class discussion).

• Step 1: Write the pertinent reactions.
• Step 2: Write the charge balance equation.
• Step 3: Write mass balance equations.
• Step 4: Write equilibrium constant expression for each chemical reaction.
• Step 5: Count the equations and unknowns. The numbers should be the same.
• Step 6: Solve those equations.

8-5 Applying the systematic treatment of equilibrium
Solubility of calcium sulfate
Step 1
Step 2
Step 3
Step 4
Step 5: counting
Step 6

Step 6