calculete ph of 10^-2 HCl and 0.001 of NaOH
Answers
Answer:
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Answer:
pH
=
11.0
Explanation:
The problem provides you with the temperature of the solution because at
25
∘
C
, which is about
298 K
, an aqueous solution has
pH + pOH = 14
−−−−−−−−−−−−−−
This means that you can express the
pH
of the solution in terms of its
pOH
, which, as you know, is defined as
pOH
=
−
log
(
[
OH
−
]
)
You can thus say that the
pH
of the solution is equal to
pH
=
14
−
pOH
pH
=
14
−
[
−
log
(
[
OH
−
]
)
]
pH
=
14
+
log
(
[
OH
−
]
)
So instead of calculating the
pH
of the solution by using the concentration of hydronium cations,
H
3
O
+
pH
=
−
log
(
[
H
3
O
+
]
)
you can do so indirectly by using the concentration of hydroxide anions.
Now, sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, which are of no interest to you here, and hydroxide anions.
Both ions are produced in
1
:
1
mole ratio with the strong base, so you can say that
[
OH
−
]
=
[
NaOH
]
=
0.001 M
Now that you know the concentration of hydroxide anions in this solution, you can say that its
pH
is equal to
pH
=
14
+
log
(
0.001
)
=
11.0
−−−−