Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.
Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric
H
3
O
+
.
Moles of nitric acid:
26.0
×
10
−
3
⋅
L
×
8.00
⋅
m
o
l
⋅
L
−
1
=
0.208
⋅
m
o
l
H
N
O
3
(
a
q
)
.
And, moles of hydrochloric acid:
88.0
×
10
−
3
⋅
L
×
5.00
⋅
m
o
l
⋅
L
−
1
=
0.440
⋅
m
o
l
H
C
l
(
a
q
)
.
This molar quantity is diluted to
1.00
L
. Concentration in moles/Litre =
(
0.208
+
0.440
)
⋅
m
o
l
1
L
=
0.648
⋅
m
o
l
⋅
L
−
1
.
Now we know that water undergoes autoprotolysis:
H
2
O
(
l
)
⇌
H
+
+
O
H
−
. This is another equilibrium reaction, and the ion product
[
H
+
]
[
O
H
−
]
=
K
w
. This constant,
K
w
=
10
−
14
at
298
K
.
So
[
H
+
]
=
0.648
⋅
m
o
l
⋅
L
−
1
;
[
O
H
−
]
=
K
w
[
H
+
]
=
10
−
14
0.648
=
?
?
p
H
=
−
log
10
[
H
+
]
=
−
log
10
(
0.648
)
=
?
?
Alternatively, we know further that
p
H
+
p
O
H
=
14
. Once you have
p
H
,
p
O
H
is easy to find. Take the antilogarithm of this to get
[
O
H
−
]
.
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