Substituting these values into the Kbexpression for aqueous NH 3 gives
Kb1.8 10 ^5
This suggests that (x3.2 10 ^3 )x. So we can approximate.
1.8 10 ^5 and x 0.57 MNH 3
The solution is 0.57 MNH 3. Our assumption that (x3.2 10 ^3 )xis justified.
You should now work Exercises 54, 56, and 58.
(3.2 10 ^3 )(3.2 10 ^3 )
x
(3.2 10 ^3 )(3.2 10 ^3 )
(x3.2 10 ^3 )
[NH 4 ][OH]
[NH 3 ]
NH 3 H 2 O NH 4 OH
initial
change
at equil (x 3.2 10 ^3 ) M 3.2 10 ^3 M
3.2 10 ^3 M 3.2 10 ^3 M
3.2 10 ^3 M
3.2 10 ^3 M
At equilibrium [OH] 3.2 10 ^3 M, so
x M 0 M ≈ 0 M
1
2
(^43)
6 5
Each Kaexpression includes [H 3 O],
so each Kaexpression must be satisfied
by the totalconcentration of H 3 Oin
the solution.
772 CHAPTER 18: Ionic Equilibria I: Acids and Bases
Problem-Solving Tip:It Is Not Always x that Can Be Neglected
Students sometimes wonder about the approximation in Example 18-15, thinking that
only xcan be neglected. We can consider neglecting one term in an expression only
when the expression involves additionor subtraction. The judgment we must make is
whether eitherof the terms is sufficiently smaller than the other that ignoring it would
not significantly affect the result. In Example 18-15, xrepresents the initialconcentra-
tion of NH 3 ; 3.2 10 ^3 represents the concentration that ionizes, which cannot be
greater than the original x. We know that NH 3 is a weakbase (Kb1.8 10 ^5 ), so only
a small amount of the original ionizes. We can safely assume that 3.2 10 ^3 x, so
(x3.2 10 ^3 )x, the approximation that we used in solving the example.
Remember that we can neverneglect a term in multiplication or division.
POLYPROTIC ACIDS
Thus far we have considered only monoproticweak acids. Acids that can furnish twoor
more hydronium ions per molecule are called polyprotic acids. The ionizations of
polyprotic acids occur stepwise, that is, one proton at a time. An ionization constant expres-
sion can be written for each step, as the following example illustrates. Consider phosphoric
acid as a typical polyprotic acid. It contains three acidic hydrogen atoms and ionizes in
three steps.
H 3 PO 4 H 2 O 34 H 3 OH 2 PO 4 Ka17.5 10 ^3
H 2 PO 4 H 2 O 34 H 3 OHPO 42 Ka26.2 10 ^8
HPO 42 H 2 O 34 H 3 OPO 43 Ka33.6 10 ^13
[H 3 O][PO 43 ]
[HPO 42 ]
[H 3 O][HPO 42 ]
[H 2 PO 4 ]
[H 3 O][H 2 PO 4 ]
[H 3 PO 4 ]
18-5