We are given:
Molarity of
C
4
H
4
N
2
O
3
=
0.100
M
Volume of
C
4
H
4
N
2
O
3
=
40
m
L
Molarity of
K
O
H
=
0.100
M
K
a
=
9.8
×
10
−
5
Chemical reaction:
C
4
H
4
N
2
O
3
+
K
O
H
→
C
4
H
3
N
2
O
3
K
+
H
2
O
Part a:
no KOH is added.
The concentration of hydrogen ion can be calculated as:
C
4
H
4
N
2
O
3
⇌
C
4
H
3
N
2
O
−
3
+
H
+
K
a
=
[
C
4
H
3
N
2
O
−
3
]
[
H
+
]
[
C
4
H
4
N
2
O
3
]
[
C
4
H
3
N
2
O
−
3
]
=
[
H
+
]
9.8
×
10
−
5
=
[
H
+
]
2
[
C
4
H
4
N
2
O
3
]
9.8
×
10
−
5
=
[
H
+
]
2
0.100
[
H
+
]
2
=
9.8
×
10
−
6
[
H
+
]
=
3.13
×
10
−
3
M
The pH value of the solution is:
p
H
=
−
log
[
H
+
]
p
H
=
−
log
(
3.13
×
10
−
3
)
p
H
=
2.50
Part b:
20 mL of KOH is added.
We will find the initial moles of the reactants:
M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100
M
×
0.040
L
=
0.0040
m
o
l
n
K
O
H
=
0.100
M
×
0.020
L
=
0.0020
m
o
l
Moles of the acid and salt after the addition of KOH:
n
C
4
H
4
N
2
O
3
=
0.00400
−
0.0020
=
0.0020
m
o
l
n
C
4
H
3
N
2
O
3
K
=
0.00200
m
o
l
The concentration of acid and salt:
M
=
n
V
V
=
40
+
20
=
60
m
L
[
C
4
H
4
N
2
O
3
]
=
0.0020
m
o
l
0.060
L
=
0.033
M
[
C
4
H
3
N
2
O
3
K
]
=
0.0020
m
o
l
0.060
L
=
0.033
M
The pH value is calculated as:
p
H
=
−
log
(
K
a
)
+
log
(
[
C
4
H
3
N
2
O
3
K
]
[
C
4
H
4
N
2
O
3
]
)
p
H
=
−
log
(
9.8
×
10
−
5
)
+
log
(
0.033
0.033
)
p
H
=
4.01
Part c:
39 mL of KOH is added.
We will find the initial moles of the reactants:
M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100
M
×
0.040
L
=
0.0040
m
o
l
n
K
O
H
=
0.100
M
×
0.039
L
=
0.0039
m
o
l
Moles of the acid and salt after the addition of KOH:
n
C
4
H
4
N
2
O
3
=
0.00400
−
0.0039
=
0.0001
m
o
l
n
C
4
H
3
N
2
O
3
K
=
0.0039
m
o
l
The concentration of acid and salt:
M
=
n
V
V
=
40
+
39
=
79
m
L
[
C
4
H
4
N
2
O
3
]
=
0.0001
m
o
l
0.079
L
=
0.00126
M
[
C
4
H
3
N
2
O
3
K
]
=
0.0039
m
o
l
0.079
L
=
0.0494
M
The pH value is calculated as:
p
H
=
−
log
(
K
a
)
+
log
(
[
C
4
H
3
N
2
O
3
K
]
[
C
4
H
4
N
2
O
3
]
)
p
H
=
−
log
(
9.8
×
10
−
5
)
+
log
(
0.0494
0.00126
)
p
H
=
4.01
+
1.59
p
H
=
5.60
Part d:
40 mL of KOH is added.
We will find the initial moles of the reactants:
M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100
M
×
0.040
L
=
0.0040
m
o
l
n
K
O
H
=
0.100
M
×
0.040
L
=
0.0040
m
o
l
Moles of the acid and salt after the addition of KOH:
n
C
4
H
4
N
2
O
3
=
0.00400
−
0.0040
=
0.0
m
o
l
n
C
4
H
3
N
2
O
3
K
=
0.0040
m
o
l
The concentration of acid and salt:
M
=
n
V
V
=
40
+
40
=
80
m
L
[
C
4
H
4
N
2
O
3
]
=
0.0
m
o
l
0.080
L
=
0.0
M
[
C
4
H
3
N
2
O
3
K
]
=
0.0040
m
o
l
0.080
L
=
0.050
M
The pH value is calculated as:
p
H
=
7
+
1
2
(
p
K
a
+
log
[
C
4
H
3
N
2
O
3
K
]
)
p
H
=
7
+
1
2
(
4.01
+
log
(
0.05
)
)
p
H
=
7
+
1
2
(
4.01
−
1.30
)
p
H
=
7
+
1.35
p
H
=
8.35
Part e:
41 mL of KOH is added.
We will find the initial moles of the reactants:
M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100
M
×
0.040
L
=
0.0040
m
o
l
n
K
O
H
=
0.100
M
×
0.041
L
=
0.0041
m
o
l
Moles of the base remains left over after the addition of KOH:
n
K
O
H
=
0.0041
−
0.0040
=
0.0001
m
o
l
The concentration of base:
M
=
n
V
V
=
40
+
41
=
81
m
L
[
K
O
H
]
=
0.0001
m
o
l
0.081
L
=
0.00123
M
The concentration of hydroxide ion is equal to 0.00123 M because it dissolves as:
K
O
H
→
K
+
+
O
H
−
The pOH value of the solution:
p
O
H
=
−
log
[
O
H
−
]
p
O
H
=
−
log
(
0.00123
)
p
O
H
=
2.91
The pH value is calculated as:
p
H
+
p
O
H
=
14
p
H
+
2.91
=
14
p
H
=
11.1