Answer:The pH of the solution is given by pH=−log([H3O+])
Explanation:so you can't use
pH
=
−
log
(
0.150
)
because that's the concentration of the hydroxide anions,
OH
−
, not of the hydronium cations,
H
3
O
+
. In essence, you calculated the
pOH
of the solution, not its
pH
.
Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a
1
:
1
mole ratio.
NaOH
(
a
q
)
→
Na
+
(
a
q
)
+
OH
−
(
a
q
)
So your solution has
[
OH
−
]
=
[
NaOH
]
=
0.150 M
Now, the
pOH
of the solution can be calculated by using
pOH
=
−
log
(
[
OH
−
]
)
−−−−−−−−−−−−−−−−−−−−
In your case, you have
pOH
=
−
log
(
0.150
)
=
0.824
Now, an aqueous solution at
25
∘
C
has
pH + pOH
=
14
−−−−−−−−−−−−−−so you can't use
pH
=
−
log
(
0.150
)
because that's the concentration of the hydroxide anions,
OH
−
, not of the hydronium cations,
H
3
O
+
. In essence, you calculated the
pOH
of the solution, not its
pH
.
Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a
1
:
1
mole ratio.
NaOH
(
a
q
)
→
Na
+
(
a
q
)
+
OH
−
(
a
q
)
So your solution has
[
OH
−
]
=
[
NaOH
]
=
0.150 M
Now, the
pOH
of the solution can be calculated by using
pOH
=
−
log
(
[
OH
−
]
)
−−−−−−−−−−−−−−−−−−−−
In your case, you have
pOH
=
−
log
(
0.150
)
=
0.824
Now, an aqueous solution at
25
∘
C
has
pH + pOH
=
14
−−−−−−−−−−−−−−
