Download Chapter 9: Aqueous Solutions and Chemical Equilibria

Chapter 9:
Aqueous Solutions and Chemical
Equilibria
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9A The chemical composition of aqueous solutions
9A-1 Classifying Solutions of Electrolytes
Electrolytes form ions when dissolved in solvent and thus
produce solutions that conduct electricity.
• Strong electrolytes ionize almost completely
• Weak electrolytes ionize only partially
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9A-2 Acids and bases
Brønsted–Lowry Theory of Acids and Bases
Acid: the substance act as proton (H+) donor
Base: the substance act as proton (H+) acceptor
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Neutralization
acid + base → salt + water ???
In aqueous phase:
• HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
• CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
• NaCN(aq) + HCl(aq) → NaCl(aq) + HCN(aq)
• NaCN(aq) + CH3COOH(aq) ⇌ HCN(aq) + CH3COONa(aq)
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9A-3 Amphiprotic species
Amphiprotic species: species that have both acidic and
basic properties.
Ex., dihydrogen phosphate ion, H2PO4-
H 2 PO4− + H 3O +
H 3 PO4 + H 2O
H 2 PO4− + OH −
HPO42− + H 2O
• zwitterion—a neutral species that has both a positive and
a negative charge. EX. glycine:
NH 2CH 2 COOH → NH 3+ CH 2COO −
• simple amino acids are an important class of amphiprotic
compounds
• water is an amphiprotic solvent, other common
amphiprotic solvents include methanol, ethanol, and
anhydrous acetic acid.
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9A-4 Autoprotolysis
Autoprotolysis (autoionization): the spontaneous reaction
of molecules of a substance to give a pair of ions.
Examples:
H 2O + H 2O
H 3O + + OH −
CH 3OH + CH 3OH
HCOOH + HCOOH
NH 3 + NH 3
CH 3OH 2 + +CH 3O −
HCOOH 2 + + HCOO −
NH 4 + + NH 2 −
• The hydronium ion (H3O+) and hydroxide ion (OH-)
concentrations in pure water at 25oC are only about 10-7 M.
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9A-5 Strengths of Acids and Bases
Strong acids and strong bases
• The common strong acids: HCl, HBr, HI, HClO4, HNO3, the
first proton in H2SO4, and the organic sulfonic acid RSO3H.
• The common strong bases: NaOH, KOH, Ba(OH)2, and the
quaternary ammonium hydroxide R4NOH, where R is an
alkyl group such as CH3 or C2H5.
Dissociation reactions and relative strengths of some
common acids and their conjugate bases.
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Leveling solvent and Differentiating solvent
• Leveling solvent: such as water, several acids are completely
dissociated and show the same acid strength.
HClO4 + H 2O → ClO4− + H 3O +
HCl + H 2O → Cl − + H 3O +
• Differentiating solvent: such as acetic acid, various acids
dissociate to different degrees and have different strength.
HClO4 + CH 3COOH
HCl + CH 3COOH
ClO4− + CH 3COOH 2+
Cl − + CH 3COOH 2+
K a1
Ka2
K a1 > K a 2
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9B Chemical equilibrium
9B-1 The Equilibrium State
A chemical equilibrium is independent of the route to the
equilibrium state.
At equilibrium, the amounts of reactants and products are
constant because the rates of the forward and reverse
processes are exactly the same.
Equilibrium is a dynamic process.
This relationship can be altered by applying stressors such
as changes in temperature, in pressure, or in total
concentration of a reactant or a product.
Le Châtelier’s principle: The position of chemical
equilibrium always shifts in a direction that tends to
relieve the effect of an applied stress.
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• The mass-action effect: A shift in the position of an
equilibrium caused by adding one of the reactants or
products to a system.
• Effect of Temperature on Equilibrium
The equilibrium constant (K) depend on temperature:
K1
∆H o 1 1
ln( ) = −
( − )
K2
R T1 T2
For endothermic reaction (△Ho > 0), increase T,
increase K, shift right for restoring the equilibrium.
For exothermic reaction (△Ho < 0), increase T, decrease
K, shift left for restoring the equilibrium.
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9B-2 Equilibrium-Constant Expressions
w moles of W react with x moles of X to form y moles of Y and z
moles of Z
wW + xX
yY + zZ
The equilibrium-constant expression:
y
z
Y ] [Z ]
[
K=
w
x
[W ] [ X ]
• molar concentrations if they represent dissolved solutes.
• partial pressures in atmospheres if they are gas-phase
reactants or products. [Z]z is replaced with pz (partial
pressure of Z in atmosphere).
• No term for Z is included in the equation if this species is a
pure solid, a pure liquid, or the solvent of a dilute solution.
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• An equilibrium-constant expression yields no information
concerning the rate of a reaction.
• Some reactions have highly favorable equilibrium constants
but are of little analytical use because they are slow.
• This limitation can often be overcome by the use of a catalyst.
• The exact equilibrium-constant expression takes the form:
aYy a Zz
K= w x
aW a X
where aY, aZ, aW, and aX are the activities of species Y, Z, W,
and X. (detailed in section 10.B)
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9B-3 Types of Equilibrium Constants in Analytical Chemistry
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Stepwise and Overall Formation constants for Complex ions
• Stepwise formation constant
• Overall (Cumulative) formation constant
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9B-4 Applying the Ion-Product Constant for Water
2H 2O
+
H 3O +OH
-
 H 3O +  OH - 
K=
2
[ H 2O ]
[H 2O]2 can be considered as constant.
K [ H 2O] = K w =  H 3O+  OH - 
K w : ion-product for water.
2
-logK w = -log  H 3O+  - log OH - 
pK w = pH + pOH
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 H 3O+ 
= OH - 
from H 2O
from H 2O
OH -  = 0.200 + OH - 
= 0.200 +  H 3O+ 
≈ 0.200 M
from H 2O
from H 2O
pOH = -log0.200 = 0.699
Kw
1x10-14
 H 3O+  =
=
= 5.00x10-14 M
0.200
OH 
pH = -log 5.00x10-14 = 13.301
Check approximation
OH -  = 0.200 + 5.00x10-14 ≈ 0.200 OK
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9B-5 Using Solubility-Product Constants
For example, when an excess of barium iodate is
equilibrated with water:
Ba(IO3 ) 2(s)
2+
Ba (aq)
+ 2IO3(aq)
K sp =  Ba 2+   IO3- 
2
K sp : solubility product constant
the position of this equilibrium is independent of the
amount of Ba(IO3)2 as long as some solid is present.
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The Solubility of a Precipitate in Pure Water
2+
Ba (aq)
+ 2IO-3(aq)
Ba(IO3 ) 2(s)
s
2s
(s)(2s) 2 = 4s3 = 1.57x10-9 = K sp
7.32x10-4 mmol Ba(IO3 ) 2(s)
mL
K sp = 1.57x10−9
s = 7.32x10-4 M
× 500 mL ×
0.487 g Ba(IO3 ) 2
mmol Ba(IO3 ) 2
= 0.178 g Ba(IO3 )
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The Effect of a Common Ion on the Solubility of a Precipitate
Ba(IO3 ) 2(s)
i
2+
Ba (aq)
+ 2IO-3(aq)
0.02
c
e
K sp = 1.57x10−9
-
s
2s
0.02 + s 2s
assume 0.02 >>s
(0.02 + s)(2s) 2 ≈ (0.02)(2s) 2 = 0.08 s2 = 1.57x10-9 = K sp
s = solubility of Ba(IO3 ) 2 = 1.40 x 10-4 M
check: (0.02 + 1.40 x 10-4 ) ≈ (0.02)
OK
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Step 1
0.01 mmol
= 2.00 mmol
mL
0.1 mmol
IO3− = 100 mL x
= 10.0 mmol
mL
2+
Ba(IO3 ) 2(s)
Ba (aq)
+ 2IO3(aq)
Ba 2+ = 200 mL x
i
2.00 mmol 10.0 mmol
c
-2.00 mmol -4.00 mmol
f
2.00 mmol
0 mmol 6.00 mmol
6.00 mmol
 IO3-  =
= 0.0200 M
200 mL + 100 mL
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step 2
2+
Ba (aq)
+ 2IO-3(aq)
Ba(IO3 ) 2(s)
i
c
s
0.02
2s
e
s
0.02 + 2s
K sp = 1.57x10−9
assume 0.02 >> 2s
(s)(0.02 + 2s) 2 ≈ (s)(0.02) 2 = 4.0x10-4s = 3.93x10-6 = K sp
s = solubility of Ba(IO3 ) 2 = 3.93x10-6 M
check: (0.02 + 2x3.93x10-6 ) ≈ (0.02)
OK
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9B-5 Using Acid/Base Dissociation Constants
When a weak acid or a weak base is dissolved in water,
partial dissociation occurs.
For a weak acid
HA + H 2O
H 3O
+
+ A
−
[H3O + ][A − ]
Ka =
[HA]
K a : acid dissociation constant
For a weak base
B + H 2O
BH
+
+ OH
−
[BH + ][OH − ]
Kb =
[B]
K b : base dissociation constant
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Dissociation Constants for Conjugate Acid/Base Pairs
NH4+ and NH3 for example:
NH 3 +H 2O
+
4
NH +H 2O
+
4
NH +OH
 NH +4  OH - 
= 1.75 x10−5
Kb =
[ NH 3 ]
-
NH 3 +H 3O
[ NH 3 ]  H 3O+ 
Ka =
= 5.70 x10−10
[ NH 4 ]
+
K a K b =  H 3O +  OH -  = K w = 1.00 x10−14
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Hydronium Ion Concentration of Solutions of Weak Acids
When the weak acid HA is dissolved in water, two
equilibria produce hydronium ions:
HA+H 2 O
2H 2 O
+
H 3O +A
H 3O + +OH -
-
 H 3O +   A - 
Ka =
[ HA ]
K w =  H 3O +   OH - 
C HA = [ HA ] +  A - 
assume
 H 3O + 
>>  H 3O + 
from acid
from wate r
 A -  ≈  H 3O + 
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M ethod 1
HA
+
+ H 2O
H 3O + A
c HA -x
x
-
 H 3O +   A - 
Ka =
[ HA ]
x
 H 3O +   A - 
[H 3O + ]2
=
Ka =
c HA − [H 3O + ]
[ HA ]
2
 H 3O +  +K a  H 3O +  -K a c HA =0
-K a + K a2 +4K a c HA
 H 3O  =
2
+
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M ethod 2
 H 3O +   A - 
[H 3O + ]2
Ka=
=
c HA -[H 3O + ]
[ HA ]
Assume c HA >>[H 3O + ]
 H 3O + 
Ka=
c HA
2
 H 3O +  = K a c HA
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Method 2
Method 1
c HA /K a ≈ 104
~0.5% error
c HA /K a ≈ 102
~5%
error
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9C Buffer solutions
Buffer solution: A solution that changes pH only slightly
when small amounts of a strong acid or a strong base are
added.
Examples:
• A weak acid and its conjugate base, EX. CH3COOH +
CH3COONa.
• A weak base and its conjugate acid, EX. NH3 + NH4Cl
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9C-1 Calculating the pH of Buffer Solutions
Example, for a solution containing a weak acid, HA, and its
conjugate base, A-, may be acidic, neutral, or basic, depending on
the positions of two competitive equilibria:
+
HA + H 2O
−
H 3O + A
−
−
A + H 2O
OH + HA
 H 3O +   A− 
Ka =
[ HA]
OH −  [ HA] K w
=
Kb =
Ka
 A− 
[ HA] = cHA −  H 3O +  + OH − 
 A−  = cNaA +  H 3O +  − OH − 
assume
cHA and cNaA >>  H 3O +  and OH − 
[ HA] ≈ cHA
therefore
 A−  ≈ cNaA
c
 H 3O +  ≈ K a HA
cNaA
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The Henderson-Hasselbalch equation
H 3O+ +A -
HA+H 2O
 H 3O+   A - 
Ka =
[ HA]
[A − ]
c
− log[H 3O ] = −logK a + log
≈ − log K a + log NaA
[HA]
cHA
+
pH ≈ pK a +log
c NaA
c HA
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The effect of dilution
The effect of dilution of the pH of buffered and unbuffered
solutions. A HA with Ka = 1x10-4 and cHA = 1.00 M for example
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The Composition of Buffer Solutions as a Function of pH: Alpha
Values
Define: cT is the sum of the analytical concentrations of acetic
acid and sodium acetate in a typical buffer solution:
cT = c HOAc + c NaOAc
α0
[ HOAc ]
=
OAc − 
α1 =
cT
cT
therefore, α 0 + α1 = 1
[OAc − ] =
K a [ HOAc ]
 H 3O + 
 H 3O +  + K a
K a [ HOAc ]

cT = [ HOAc ] + OAc  = [ HOAc ] +
)
= [ HOAc ] (
 H 3O + 
 H 3O + 
−
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Rearranged equation
[ HOAc ] =
cT
α0
 H 3O + 
 H 3O +  + K a
[ HOAc ] =
=
cT
 H 3O + 
 H 3O +  + K a
OAc − 
Ka
=
α1 =
cT
 H 3O +  + K a
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Figure 9-5 Graph shows the variation in a with pH.
Note that most of the transition between a0 and a1 occurs within ±1 pH unit
of the crossover point of the two curves.
The crossover point where α0 = α1 = 0.5 occurs when pH = pKHOAc = 4.74.
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Buffer capacity
• Define buffer capacity: The ability of how well a solution
resists changes in pH when acid or base is added.
• Quantitative equation of buffer capacity: The number of
moles of a strong acid or a strong base that causes 1.00 L of
the buffer to undergo a 1.00 unit change in pH.
dc
dc
β = b =− a
Figure 9-6 Buffer
dpH
dpH
capacity as a
dcb is the number
of moles per liter of
strong base, and
dca is the number
of moles per liter of
strong acid added
to the buffer.
function of the
logarithm of the
ratio cNaA/cHA.
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Preparation of Buffers
• Making a buffer solution with a desired pH
[A − ]
pH = pK a + log
[ HA ]
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• Six methods for preparing buffer solutions
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End of Chapter 09
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