100 ml of 1 M Ca(OH), solution is mixed with
10 ml of 20 N H2SO4. The resulting solution is
(1) Acidic
(2) Neutral
(3) Weakly acidic (4) Basic
Answers
Answer:
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
∣
∣