Quantitative Chemical Analysis 7e

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Harris Quantitative Chemical Analysis, Eight
Edition
CHAPTER 07 ACTIVITY AND THE SYSTEMATIC
TREATMENT OF EQUILIBRIUM
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Ionic and hydrated radii of several ions
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Water Binding to Ions
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Equilibrium Constants with Concentrations and
Activities
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7-1 The Effect of Ionic Strength on Solubility of
Salts
CaSO4(s) Ca2 SO42- Ksp 2.4 X 10-5
(8-2)
When we add salt to a solution, we say that the
ionic strength of the solution increases.
We call this region the ionic atmosphere (Figure
8-2). The greater the ionic strength of a
solution, the higher the charge in the ionic
atmosphere. Each ion-plus-atmosphere contains
less net charge and there is less attraction
between any particular cation and anion.
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The effect is to reduce their tendency to come
together, thereby increasing the solubility of
CaSO4.
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Ionic strength, µ, is a measure of the total
concentration of ions in solution.
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Box7-1 Salts with Ions of Charge 2 Do Not
Fully Dissociate
Ion pair formation constant Mn(aq) Lm-(aq)
MnLm-(aq)
Ion pair
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7-2 Activity Coefficients
To account For the effect of ionic strength,
concentrations are replaced by activities
The activity of species C is its concentration
multiplied by its activity coefficients.
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Ksp ACa2ASO42- Ca2?Ca2SO42-?SO42-
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The ionic atmosphere model leads to the extended
Debye-Hückel equation, relating activity
coefficients to ionic strength
To find activity coefficients for ionic strengths
above 0.1 M (up to molalities of 2-6 mol/kg for
many salts), more complicated Pitzer equations
are usually used.
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In linear interpolation, we assume that values
between two entries of a table lie on a straight
line.
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7-3 pH Revisited
pH -logAH -logH?H
(8-8)
When we measure pH with a pH meter, we are
measuring the negative logarithm of the hydrogen
ion activity, not its concentration.
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However, the concentration of H in 0.10 M KCl
(1.26 X 10-7 M) is 26 greater than the
concentration of H in pure water (1.00 X 10-7 M).
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7-4 Systematic Treatment of Equilibrium
The systematic treatment of equilibrium is a way
to deal with all types of chemical equilibria,
regardless of their complexity. The charge
balance is an algebraic statement of
electroneutrality The sum of the positive
charges in solution equals the sum of the
negative charges in solution.
H K OH- H2PO4- 2HPO42-
3PO43- (8-11)
The coefficient in front of each species always
equals the magnitude of the charge on the ion.
H 5.1 X 10-12 M H2PO4- 1.3
X 10-6 M K 0.0550 M
HPO42- 0.0220 M OH- 0.0020 M
PO43- 0.0030 M
H K OH- H2PO4-
2HPO42- 3PO43- 5.1 X 10-12 0.0550
0.0020 1.3 X 10-6 2(0.0220) 3(0.0030)
0.0550 M 0.0550 M
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Where C is the concentration of a cation, n is
the charge of the cation, A is the
concentration of an anion, and m is the magnitude
of the charge of the anion.
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The mass balance, also called the material
balance, is a statement of the conservation of
matter. The mass balance states that the quantity
of all species in a solution containing a
particular atom (or group of atoms) must equal
the amount of that atom (or group) delivered to
the solution.
CH3CO2H CH3CO2- H Acetic acid
Acetate Mass balance for 0.050M
CH3CO2H CH3CO2- Acetic acid in water
What we put into Undissociated Dissociated
the
solution product
product 0.0250 M H3PO4 H2PO4-
HPO42- PO43-
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Box 7-2 Calcium Carbonate Mass Balance in Rivers
CaCO3(s) CO2(aq) H2O Ca2 2HCO3-
(A) Calcite
Bicarbonate
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7-5 Applying the Systematic Treatment of
Equilibrium A simple Example Ionization of Water
Step 1 Pertinent reactions. The only one is
Reaction 8-13. Step 2 Charge balance. The only
ions are H and OH-, so the charge balance is
H OH-
(8-14)
Step 3 Mass balance. Reaction 8-13 creates one H
for each OH-. The mass balance is simply H
OH-, which is the same as the charge balance
for this system. Step 4 Equilibrium constant
expression.
KW H?HOH-?OH- 1.0 X 10-14
(8-15)
This is the only step in which activity
coefficients enter the problem.
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Step 5 count equations and unknowns. We have two
equations, 8-14 and 8-15, and two unknowns, H
and OH-. Step 6 Solve.
H?HOH-?OH- 1.0 X 10-14 H 1 H
1 1.0 X 10-14 H
1.0 X 10-7 M pH -logAH -logH?H
-log(1.0 X 10-7)(1) 7.00
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Solubility of Calcium Sulfate
Step 1 Pertinent reactions. Even in such a simple
system, there are quite a few reactions
There is no way you can be expected to come up
with all of these reactions, so you will be given
help with this step. Step 2 Charge balance.
Equating positive and negative charges gives
2Ca2 CaOH H 2SO42- HSO4-
OH- (8-21)
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Step 3 Mass balance. Reaction 8-16 produces 1mole
of sulfate for each mole of calcium. No matter
what happens to these ions, the total
concentration of all species with sulfate must
equal the total concentration of all species with
calcium
Total calcium total sulfate Ca2
CaSO4(aq) CaOH SO42- HSO4-
CaSO4(aq) (8-22)
Step 4 Equilibrium constant expressions. There is
one for each chemical reaction.
Step 4 is the only one where activity
coefficients come in.
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Step 5 Count equations and unknowns. There are
seven equations (8-21 through 8-27) and seven
unknowns Ca2, SO42-, CaSO4(aq), CaOH,
HSO4-, H, and OH-. In principle, we have
all the information necessary to solve the
problem. Step 6 Solve. Well, this is not easy!
We dont know the ionic strength, so we cannot
evaluate activity coefficients. Also, where do we
start when there are seven unknowns?
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Solubility of Magnesium Hydroxide
Step 1 Pertinent reactions are listed
above. Step 2 Charge balance 2Mg2 MgOH
H OH- Step 3 Mass balance. This is a
little tricky. From Reaction 8-30, we could say
that the concentrations of all species containing
OH- equal two times the concentrations of all
magnesium species. However, Reaction 8-32 also
creates 1 OH- for each H. The mass balance
accounts for both sources of OH-
After all this work, Equation 8-34 is equivalent
to Equation 8-33.
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Step 4 Equilibrium constant expressions are in
Equations 8-30 through 8-32. Step 5 Count
equations and unknowns. We have four equations
(8-30 to 8-33) and four unknowns Mg2,
MgOH, H, and OH-. Step 6 Solve.
2Mg2 MgOH OH- (8-35)
2Mg2 K1Mg2OH- OH-