Chemistry Reference
In-Depth Information
C
2
H
3
O
2
Sodium acetate is a salt, so initially, the concentration of
is 0.195 M.
The acetate ion affects the position of the acid equilibrium.
H
2
O(
/
) E C
2
H
3
O
2
(aq)
H
3
O
(aq)
HC
2
H
3
O
2
(aq)
Initial
0.150
0.195
0
Change
Equilibrium
x
x
x
0.150
x
0.195
x
x
[C
2
H
3
O
2
][H
3
O
]
[HC
2
H
3
O
2
]
K
a
Neglecting
x
when added to or subtracted from larger quantities and using the
value of
K
a
in Table 19.2 yields:
(0.195) (
x
)
0.150
10
5
K
a
1.76
10
5
x
1.35
4
The approximations made by ignoring
x
when added to 0.195 M and subtracted
from 0.150 M are valid. Because
[H
3
O
]
10
5
M,
is
1.35
4
pH
4.868
Practice Problem 19.17
Calculate the pH of a solution containing
0.120 M
NH
3
and 0.160 M NH
4
Cl.
If strong acid or strong base is added to a buffer solution, a net reaction
takes place. To calculate the pH of such solutions, we first assume that the strong
acid or base reacts as completely as possible, given that some reactant is pres-
ent in limiting quantity, as discussed in Section 10.4. Only then do we concen-
trate on the equilibrium calculation.
EXAMPLE 19.18
Calculate the pH after 0.0300 mol of solid NaOH is added to 1.000 L of a solu-
tion containing 0.150 mol of
HC
2
H
3
O
2
and 0.195 mol of
NaC
2
H
3
O
2
.
Assume
that the volume of the solution does not change.
Solution
When a base is added to an acid, they react to form a salt (Chapter 8). Here,
the base (NaOH) is in limiting quantity because a greater number of moles of
the acid is present, and the acid and base react in a 1 : 1 ratio:
HC
2
H
3
O
2
(aq)
NaOH(s) £ NaC
2
H
3
O
2
(aq)
H
2
O(
/
)
Initial
concentrations
0.150
0.0300
0.195
Excess
Changes due to
acid-base
reaction
Concentrations
before
equilibrium
0.0300
0.0300
0.0300
0.0300
0.120
0.0000
0.225
Excess
