Tài liệu Chapter 6: Chemical Equilibrium ppt

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c. Autoprotolysis: when amphiprotic
solvents undergo self-ionization
2 H
2
O <=====> H
3
O
+
+ OH
-
K
w
This expression is called the ion-product of water and is
denoted
K
w
= [H
3
O
+
][OH
-
]
K
w
= 1.01 x 10
-14
at 25°C
pK
w
= -log K
w
= -log 1.01 x 10
-14
= 14
Acid and Base Strengths
Acid and base strengths may be determined by
using
¾ a differentiating solvent
.
That is, selecting a solvent which will accept
(or donate) a proton to one acid (or base) but
not another (example: use of glacial acetic acid
as solvent to compare HClO
4
and HCl acid
strength).
¾ a leveling solvent
is one in which all acids or
bases are dissociated to the same degree and
have equal acid (or base) strengths.
¾a differentiating solvent
CH
3
COOH + HClO
4
CH
3
COOH
2
+
+ClO
4
-
CH
3
COOH + HCl CH
3
COOH
2
+
+Cl
-
¾ a leveling solvent (H
2
O)
HClO
4
+ H
2
O H
3
O
+
+ClO
4
-
HCl + H
2
O H
3
O
+
+Cl
-
<====>
<====>
Table: Acids and Bases
Any Questions?
II. Chemical Equilibrium:
The ratio of the molar concentrations of reactants and
products is a constant at certain temperature.
H
2
O + HCOOH <===> H
3
O
+
+ HCOO
-
The square bracketed terms mean:
¾ Molar concentration if the species is a dissolved
solute
¾ Partial pressure in atm if species is a gas
¾ Unity if species is a pure liquid, pure solid, or
pure solvent (solvent in an extremely dilute
solution).
]HCOOH][OH[
]HCOO][OH[
K
2
-
3
a
+
=
III. Types of Equilibrium Constants (See Table 6-1)
A. The ion-product for water.
Water is poorly dissociated, but does undergo
autoprotolysis:
2 H
2
O <===> H
3
O
+
+ OH
-
and,
recalling that molar concentrations = 1 for
dilute solutions
K
w
= [H
3
O
+
] [OH
-
] = 1.01 X 10
-14
at 25
o
C
Also since -log K
w
= -log{[H
3
O
+
][OH
-
]}
= -log [H
3
O
+
]-log[OH
-
]
pK
w
= pH + pOH or 14 = pH + pOH
K
w
is dependent on temperature
B. Solubility Product Constants
For the solubility of Fe(OH)
3
, we can write
Fe(OH)
3
<===> Fe
3+
(aq) + 3 OH
-
(aq)
again, recalling that the molar concentration
of a pure solid = 1
we can write the solubility product:
[Fe
+3
] [OH
-
]
3
= K
sp
= 4 x 10
-38
Any Questions?