Weak Base Equilibrium continued

1

• Weak Base Equilibrium (continued)

• Every weak acid has a corresponding weak
  conjugate base.

eg

• Here, HA is a weak acid, and A- must be a weak
  base. (it also happens to be the conjugate base
  of HA)

eg

• If HCN is a weak acid, then CN- must be a weak
  base.

• As weak acids get weaker (going down the KA
  table), their corresponding conjugate bases get
  stronger.

2
eg

• H3PO4 is a weak acid, thus, H2PO4- is a weak
  base.

But

• H2PO4- could also act as a weak acid (it still
  has H left), it is an amphiprotic substance.

• A substance is amphiprotic if it has hydrogen to
  donate and also has the capacity to accept it.

eg
acid
base
eg
H2CO3
HCO3-
CO32-
acid only
amphiprotic
base only
3

• Relationships between KA KB for any conjugate
  acid base pair (eg HA A-)

generic acid
KA
conjugate base
KB
4

• Multiply KA by KB

KA x KB
H3OOH-
KA x KB KW
(where KA by KB are for a conjugate acid base
pair)
ie
Given KA for an acid, you can calculate KB for
its conjugate base. (_at_ 25?C)
KW 1.0 x 10-14
5
Calculate KB for each of the following
eg
a) HCO3-
KA x KB KW
2.3 x 10-8
KB

b) HPO42-
1.6 x 10-7
KB

6
Calculate the pH of a 1.0M solution of H2PO4-.
eg
First we need to know if H2PO4- acts as an acid
or a base.
(KA)
or
(KB)
Compare KA with KB
KA 6.2x10-8
1.3 x 10-12

KB
Thus, although H2PO4- is an amphiprotic substance
we see that KAgtKB and it is more likely to act as
an acid.
7
Finally, we use
KA
6.2x10-8
(assuming x is negligible)
x 2.5x10-4
H3O 2.5x10-4
pH 3.60