re .
Titration reveals that 12.8 ml of
5M sulphuic
Acid
required to
neutralize the sodium
hydroxide which
is taken 35 mL
in conical
flask.
What is
the molarity of the base?
Answers
Answer:
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Answer:
9.6
Explanation:
You're titrating hydrochloric acid,
HCl
, a strong acid, with sodium hydroxide,
NaOH
, a strong base, so right from the start you should know that the pH at equivalence point must be equal to
7
.
Hydrochloric acid and sodium hydroxide react in a
1
:
1
mole ratio to form water and aqueous sodium chloride
HCl
(aq]
+
NOH
(aq]
→
NaCl
(aq]
+
H
2
O
(l]
The net ionic equation for this reaction looks like this
H
3
O
+
(aq]
+
OH
−
(aq]
→
2
H
2
O
(l]
Now, the equivalence point corresponds to a complete neutralization, i.e. when you add enough strong base to completely consume all the acid present in the solution.
The
1
:
1
mole ratio tells you that at the equivalence point, the solution must contain equal numbers of moles of strong acid and strong base.
Your tool of choice here will the equation
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
pH
=
−
log
(
[
H
3
O
+
]
)
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
Now, it's very important to realize that the volume of the solution will increase as you add the strong base solution. Keep this in mind when calculating the molarity of the hydronium ions.
So, let's start calculating the corresponding pH
1
.
Before NaOH is added
−−−−−−−−−−−−−−−−−−−
Since hydrochloric acid is a strong acid, it dissociates completely in aqueous solution to form hydronium cations and chloride anions. More specifically, you have
[
H
3
O
+
]
=
[
HCl
]
=
0.1 M
This means that the pH of the solution before any strong base is added will be equal to
pH
=
−
log
(
0.1
)
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯
a
a
1
a
a
∣
∣
−−−−−−
2
.
After 50 mL of NaOH are added
−−−−−−−−−−−−−−−−−−−−−−−−−−−
So, use the definition of molarity to determine how many moles of hydronium ions, i.e. hydrochloric acid, you have in the initial solution
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
c
=
n
solute
V
soluteion
⇒
n
solute
=
c
⋅
V
solution
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
n
H
3
O
+
=
0.1 mol
L
−
1
⋅
100
⋅
10
−
3
L
=
0.010 moles H
3
O
+
Calculate how many moles of hydroxide anions, i.e. sodium hydroxide, are delivered to the solution by adding
50 mL
of
0.1-M
sodium hydroxide solution
n
O
H
−
=
0.1 mol
L
−
1
⋅
50
⋅
10
−
3
L
=
0.0050 moles OH
−
The hydroxide anions will be completely consumed by the reaction, leaving behind
n
O
H
−
=
0
→
completely consumed
n
H
3
O
+
=
0.010 moles
−
0.0050 moles
=
0.0050 moles H
3
O
+
The total volume of the solution will be
V
total
=
100 mL
+
50 mL
=
150 mL
The concentration of hydronium ions will be
[
H
3
O
+
]
=
0.0050 moles
150
⋅
10
−
3
L
=
0.0333 M
The pH of the solution will be
pH
=
−
log
(
0.0333
)
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
1.48
a
a
∣
∣
−−−−−−−−
3
.
After 98 mL of NaOH are added
−−−−−−−−−−−−−−−−−−−−−−−−−−−
This corresponds to adding an additional
48 mL
of strong base, which is equivalent to
n
O
H
−
=
0.1 mol
L
−
1
⋅
48
⋅
10
−
3
L
=
0.0048 moles OH
−
Once again, the hydroxide anions will be completely consumed by the reaction, leaving you with
n
O
H
−
=
0
→
completely consumed
n
H
3
O
+
=
0.0050
−
0.0048
=
0.00020 moles H
3
O
+
The total volume of the solution will be
V
total
=
150 mL
+
48 mL
=
198 mL
The concentration of hydronium ions will be
[
H
3
O
+
]
=
0.00020 moles
198
⋅
10
−
3
L
=
0.001010 M
The pH of the solution will be
pH
=
−
log
(
0.001010
)
≈
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯
a
a
3
a
a
∣
∣
−−−−−−
4
.
After 99.9 mL of NaOH are added
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
This corresponds to adding an additional
1.9 mL
of strong base, which is equivalent to
n
O
H
−
=
0.1 mol
L
−
1
⋅
1.9
⋅
10
−
3
L
=
0.00019 moles OH
−
The number of moles of hydroxide anions is still smaller than the number of moles of hydronium cations, so the hydroxide anions will once again by completely consumed here.
The resulting solution will contain
n
O
H
−
=
0
→
completely consumed
n
H
3
O
+
=
0.00020
−
0.00019
=
0.000010 moles H
3
O
+
The *total volume of the solution will be
V
total
=
198 mL
+
1.9 mL
=
199.9 mL
The concentration of hydronium cations will be
[
H
3
O
+
]
=
0.000010 moles
199.9
⋅
10
−
3
L
=
0.00005 M
The pH of the solution will be
pH
=
−
log
(
0.00005
)
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
4.3
a
a
∣
∣
−−−−−−−
5
.
After 100 mL of NaOH are added
−−−−−−−−−−−−−−−−−−−−−−−−−−−−
→
EQUIVALENCE POINT
Adding an additional
0.1 mL
of strong base to the solution will correspond to a complete neutralization.
All the moles of hydronium cations that were initially present in the solution are now completely consumed. The resulting solution is neutral, since it only contains water and aqueous sodium chloride.
The concentration of hydronium cations corresponds to that of pure water at room temperature,
1.0
⋅
10
−
7
M
. The pH of the solution is
pH
=
−
log
(
1.0
⋅
10
−
7
)
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯
a
a
7
a
a
∣
∣
−−−−−−
6
.
After 100.1 mL of NaOH are added
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
This corresponds to adding an additional
0.1 mL
of strong base to a neutral solution, so you can expect the pH to be higher than
7
.
You will have
n
O
H
−
=
0.1 mol
L
−
1
⋅
0.1
⋅
10
−
3
L
=
0.00001 moles OH
−
The total volume of the solution will be
V
total
=
200 mL
+
0.1 mL
=
201 mL
The concentration of hydroxide anions will be
[
OH
−
]
=
0.000010 moles
200.1
⋅
10
−
3
L
=
0.000049975 M
The pOH of the solution will be
pOH
=
−
log
(
0.000049975
)
≈
4.3
Since you know that, at room temperature
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
pH
+
pOH
=
14
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−
you can say that the pH of the solution will be
pH
=
14
−
4.3
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
9.7
a
a
∣
∣
−−−−−−−