Chemistry Reference
In-Depth Information
The equilibrium constant expression and the value of
K
b
from Table 19.2 are
used to solve for
x:
[OH
][NH
4
]
[NH
3
]
x
2
0.165
10
5
K
b
x
1.77
Ignoring
x
when subtracted from 0.165 yields
x
2
10
6
2.92
1
Taking the square root of both sides of the equation yields
10
3
x
1.70
9
10
3
M.
The hydroxide ion concentration is
1.70
9
As a check, we can see
10
3
M
that
is 1% of 0.165 M, so the approximation is acceptable.
We can now use the
1.70
9
K
w
expression to calculate the hydronium ion
concentration:
[H
3
O
][OH
]
10
14
K
w
1.0
[H
3
O
] (1.70
9
10
3
)
10
14
1.0
Thus,
[H
3
O
]
10
12
M
5.85
1
and
10
12
)
pH
log
(5.85
1
11.233
Practice Problem 19.14
Calculate the pH of a 0.180 M
solution.
HC
2
H
3
O
2
Snapshot Review
❒
There is
some
hydronium ion and
some
hydroxide ion in every dilute
aqueous solution, whether acidic, basic, or neutral.
❒
The pH is the negative logarithm of the
hydronium ion
concentration,
not of the concentration of any other ion.
ChemSkill Builder 18.2
10
3
M
H
3
O
10
3
M
OH
.
A. Calculate the pH of (a) and (b)
B. Calculate the pH of the solution of Snapshot Review Problem A,
Section 19.2.
1.00
1.00
19.4
Buffer Solutions
An aqueous solution of a weak acid or a weak base contains two substances
that react with each other to some extent—the acid or base and the water. When
another substance is added to this solution, it can affect the original pair of reac-
tants without necessarily reacting directly with either. It may, as predicted by
LeChâtelier's principle, suppress the reaction of the original two reactants. For
example, if sodium acetate,
NaC
2
H
3
O
2
,
is added to an aqueous solution of acetic
acid,
HC
2
H
3
O
2
,
it does not react directly with the acid or with the water. Instead,
