Eight equations you will always be given the k a or k

eight equations, you will always be given the K
a
or K
b
.
Some K's that are worth memorizing are K
a
for acetic acid =
1.8 x 10
-5
, and K
b
for NH
3
= 1.8 x 10
-5
. The fact that they are the same is a coincidence. You will never be given the K's of the
other six equations. You'll need to calculate them.
Let's look at the third equation:
A
-
+ H
2
O
←
→
HA + OH
-
This is a reaction of a conjugate base with water (notice that it is uncharged water; if it was charged water,
such as H
3
O
+
, then we could not call it K
cb
). Let's call it K
cb
. There is an equation for
calculating K
cb
. K
w
= K
a
x K
cb
. Therefore, K
cb
= (K
w
/K
a
). K
w
is the hydrolysis constant for water at
25
o
C. It is 1 x 10
-14
. You will always be given K
a
(or K
b
) if you need that for a calculation. K
cb
= (1 x
10
-14
)/(1.8 x 10
-5
) = 5.56 x 10
-10
.
Let's look at the second equation:
A
-
+ H
3
O
+
→ HA + H
2
O
This is a reaction of a conjugate base with a strong acid. (Note that each of the seven strong acids react
with H
2
O to form H
3
O
+
, so H
3
O
+
becomes the new strong acid in water.) There is no name for the K for
this reaction. Let's give it a name. Let's call it K
cbsa
(K
conjugatebasestrongacid
). What is the equilibrium
constant for this reaction? In looking at the right side of the reaction note that the right side of the
equation is a weak acid + water. Hence, this reaction is the reverse of the first reaction, weak acid +
water. Since the reaction is reversed, then the K must be inverted. Hence, K
cbsa
= 1/K
a
= 1/(1.8 x 10
-5
) =
5.6 x 10
4
. Note that reactions with strong acids (or strong bases) have large K's, and reactions with weak
acids (or weak bases) have small K's.
Let's look at the fourth equation:
HA + OH
-
→ H
2
O + A
-
This is a reaction of a weak acid and a strong base. Let's call the K for this reaction K
wasb
=
K
weakacidstrongbase
. What is the K for this reaction? Look on the right side of this reaction. Looks familiar?
That's the reaction of a conjugate base with water. Hence, this equation is the reverse of the third
equation. Since the third equation is reversed, then the K needs to be inverted. So the K for this equation
= K
wasb
= 1/K
cb
= 1/(K
w
/K
a
) = K
a
/K
w
= (1.8x10
-5
)/(1x10
-14
) = 1.8 x 10
9
. Note that a reaction with a strong
base has a large K.

Chem 162-2016 Chapter 17 Tro Acid-Base Equilibria 54
SEVEN STRONG ACIDS
transparency
ET: Discuss strong acids and strong bases and the strengths of their conjugate bases and acids; practice both.
Very weak conj. base
(i.e., not a base but a
K
a
Acid spectator ion)*
~10
6
HCl + H
2
O H
3
O
+
+ Cl
-
K
cb
= ~10
-23
~10
8
HBr + H
2
O H
3
O
+
+ Br
-
K
cb
= ~10
-22
~10
9
HI + H
2
O H
3
O
+
+ I
-
K
cb
= ~10
-20
~10
3
H
2
SO
4
+ H
2
O H
3
O
+
+ HSO
4
-**
K
cb
= ~10
-17
~2x10
1
HNO
3
+ H
2
O H
3
O
+
+ NO
3
-
K
cb
= ~10
-15
~10
10
HClO
4
+ H
2
O H
3
O
+
+ ClO
4
-
K
cb
= ~10
-24
~10
1
HClO
3
+ H
2
O H
3
O
+
+ ClO
3
-
K
cb
= ~10
-15
*These conjugate bases are known as “nominal bases”, i.e., a base by definition, but effectively not a base.
**Although HSO
4
-
is effectively not a base, it is effective an acid. HSO
4
-
+ H
2
O ←
→
H
3
O
+
+ SO
4
2-
All other acids are weak acids, e.g., HA (acetic acid), H
2
SO
3
, RNH
3
+
, BF
3
CH
3
COOH + H
2
O H
3
O
+
+ CH
3
COO
-
Note inverse relationship between the strength of the acid and the strength of the conjugate base. A strong acid
(e.g., HCl) has a very weak conjugate base (Cl