Calculate the pH of 1 x 10-3 mol/L solution of NaOH in water.
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Answer:
Explanation:We examine the equilibrium:
2
H
2
O
(
l
)
⇌
H
3
O
+
+
−
O
H
Under standard conditions,
[
H
3
O
+
]
[
H
O
−
]
=
10
−
14
.
And if we take
log
10
of both sides:
log
10
[
H
3
O
+
]
+
log
10
[
H
O
−
]
=
−
14
Om rearrangement:
14
=
−
log
10
[
H
3
O
+
]
−
log
10
[
H
O
−
]
But by definition,
−
log
10
[
H
3
O
+
]
=
p
H
, and
log
10
[
H
O
−
]
=
p
O
H
.
And thus our definining relationship:
p
H
+
p
O
H
=
14
.
Since (finally!)
p
O
H
=
−
log
10
[
H
O
−
]
=
−
log
10
(
1
×
10
−
3
)
=
−
(
−
3
)
=
3
.
And if
p
O
H
=
3
,
p
H
=
14
−
3
=
?
?
In A level, you do have to remember the result:
p
H
+
p
O
H
=
14
. With this is mind, these problems become trivial.
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