Transcript Slide 1
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 BaF2 with NaF
F- is the “common ion”
(B) No common ion: “inert electrolyte effect”
or “diverse ion effect”: Add Na2SO4 to saturated
solution of AgCl Increases solubility of AgCl
8-1 The effect of ionic strength on
solubility of salts
• 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.
What is ionic strength?
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Ionic strength, μ, is a measure of the total concentration of ions in solution.
m = ½SCiZi2
where Ci is the concentration of the ion, actually a ratio of Ci/1 M, Zi
= charge on each individual ion.
Example: Find the ionic strength of (a) 0.10 M NaCl; (b) 0.020 M KBr plus 0.01
M Na2SO4.
Solution:
(a) the soultion contains 0.10 M Na+ and 0.10 M Clm = 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
SO42-.
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.
8-2 Activity Coefficients
• To account for the effect of ionic strength, concentrations
in the calculation of equilibrium constant show be
replaced by activities: ẲC = [C]*γC
• The activity of species j is its concentration multiplied by
its activity coefficient, ai = Ciƒi
ƒi = activity coefficient.
• At low ionic strength, activity coefficients approach unity.
Activity and Activity
Coefficients
• Calculation of Activity Coefficients
• Extended Debye-Huckel Equation:
• 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)
• 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
Activity coefficient under high ionic
strength
Activity coefficient for non-ionic
compounds
• 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.
Diverse Ion (Inert Electrolyte)
Effect:
• Kspo = aAg+ . aCl- = 1.75 x 10-10
• Adding Na2SO4 to saturated solution of AgCl, at
high concentration of diverse (inert) electrolyte:
higher ionic strength, m
• aAg < [Ag ] ; aCl < [Cl ]
• aAg . aCl < [Ag ] [Cl ]
+
+
+
-
-
-
+
-
• Kspo < [Ag+] [Cl-] ; Kspo < [Ag+] = solubility
• Solubility = [Ag+] > Kspo
Equilibrium calculations using
activities
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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
Kspo = (aPb)1(aI)2 = ([Pb2+]Pb )1([I-]I )2
Kspo = ([Pb2+] [I-]2)(Pb I2 ) = Ksp (Pb I2 )
Ksp = Kspo / (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
• The solubility is increased by approx. 43%
8-3 pH revisited
Addition example: Calculate the pH of water containing 0.010 M KCl at 25 oC.
(see in class discussion).
8-4 Systematic treatment of
equilibrium
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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
Solubility of magnesium hydroxide
Step 6