Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
1
CHEMISTRY 1622016
CHAPTER 18
ACIDBASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
PROF. E. TAVSS
Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
2
CHEMICAL, ACID AND BASE EQUILIBRIA
transp
pH
[H
+
]
14
Basic
1x10
14
H
2
O + H
2
O
←
→
H
3
O
+
+ OH

K
w
= 1x10
14
7
Neutral
K
w
= [H
+
][OH

] = 1 x 10
14
pH + pOH = 14
K
w
= K
a
x K
cb
= K
ca
x K
b
= 1 x 10
14
*
*a & b = acid and base;
ca & cb = conjugate acid and conjugate base
pK
a
+ pK
b
= 14
0
Acidic
1x10
o
A pH scale of 1 to 13 is common, but pH’s of 1 to
+15 are possible.
pH = log[H
+
]
pH
[H
+
]
[H
+
] = 10
pH
pOH = log[OH

]
[OH

] = 10
pOH
pOH
[OH

]
HendersonHasselbalch equation: pH = pK
a
+ log([B]/[A])
where acid = acid or conjugate acid, and base = base or conjugate base.
Also, pOH = pK
base
+ log([Acid]/[Base])
pH + pOH = 14
pH + pOH = 14
[H
+
][OH

] = 1 x 10
14
[H
+
][OH

] = 1 x 10
14
pOH = log[OH

]
[OH

] = 10
pOH
pH = log[H
+
]
[H
+
] = 10
pH
Chem 1622016 Chapter 18 Acidbase & solubility equilibria lecture notes
3
MATHEMATICAL TRICKS FOR AVOIDING QUADRATIC EQUATIONS
RESULTING FROM SOLVING ICE TABLE PROBLEMS
(1) Small K rule.
page 33.
2CO
2
(g)
←
→
2CO(g) + O
2
(g)
If the equilibrium constant is small, such as
≤
10
3
K, then this means that there is very little
substance on the right side of the equation (which is usually the product side) and virtually all of
the substance is on the left side (which is usually the reactant side).
Therefore, drop the X in the
ICE table when the X is in a format such as YX on the reactant side, or Y+X on the product side,
because no significant amount of new product will be forming.
The requirement for using the
small K rule is that the error caused by this approximation must be < 5% of the value of Y.
(2) Large K rule.
page 34.
2NO(g) + Cl
2
(g)
←
→
2NOCl(g)
If the equilibrium constant is large, such as
≥
10
3
, this means that almost all of the substances are
on the product side.
Handle this ICE table by first finding the limiting reactant and then bringing
the reaction to completion.
This will result in the concentration of the limiting reactant becoming
zero and the concentrations of the product becoming large.
(There will be some times when a zero
concentration is obtained for a substance, that we are required to have a finite value for that
substance.
In that case we would have to bring the reaction slightly back to the left, i.e., back to a
finite equilibrium value, using the “RighttoLeft K Rule.)
(3) Righttoleft K Rule.
After using the large K rule and bringing the reaction to completion, sometimes it’s necessary to
go from right to left (hence, “RighttoLeft K Rule) in order to provide a finite value of a
substance having “0” concentration on the left side of the equilibrium equation.
When going from
right to left, realize that K is large, so there is a large quantity on the right side and a small
quantity on the left side of the equation.
Therefore, to avoid a quadratic equation, one may drop
the “X” from a larger number on the right side (X would be very small in going from right to left
when K is large) and one may correspondingly drop the “+X” from a larger number on the left
side.
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 Summer '09
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 Chemistry, pH, Solubility, solubility equilibria lecture, equilibria lecture notes