What volume of 0.841 M KOH solution is required to make 3.53 L of a solution with pH of 12.80?
Answers
Answer:
Explanation:
Start by using the
pOH
of the solution to figure out the concentration of hydroxide anions,
OH
−
. You should know that
pOH
=
−
log
(
[
OH
−
]
)
−−−−−−−−−−−−−−−−−−−−
To solve for the concentration of hydroxide anions, rearrange the equation as
log
(
[
OH
−
]
)
=
−
pOH
This can be written as
10
log
(
[
OH
−
]
)
=
10
−
pOH
which is equivalent to
[
OH
−
]
=
10
−
pOH
Now, an aqueous solution at room temperature has
pH
+
pOH
=
14
−−−−−−−−−−−−−−−−
This means that the
pOH
of the target solution is equal to
pOH
=
14
−
12.4
=
1.6
Consequently, the target solution must have
[
OH
−
]
=
10
−
1.6
=
2.51
⋅
10
−
2
M
As you know, molarity is defined as the number of moles of solute present in
1 L
of solution. This means that the number of moles of potassium hydroxide needed to ensure that concentration of hydroxide anions is equal to
3.55
L solution
⋅
2.51
⋅
10
−
2
moles OH
−
1
L solution
=
8.91
⋅
10
−
2
moles OH
−
All you have to do now is figure out what volume of the stock solution contains the needed number of moles of hydroxide anions
8.91
⋅
10
−
2
moles OH
−
⋅
1 L solution
0.855
moles OH
−
=
0.104 L
Expressed in milliliters, the answer will be
V
stock KOH
=
104 mL
−−−−−−−−−−−−−−−−−
I'll leave the answer rounded to three sig figs.
So, in order to prepare this solution, take
104 mL
of
0.855 M
stock potassium hydroxide solution and add enough water until the final volume of the solution is equal to
3.55 L
.