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Calculate the pH at the following points in a titration of 40 mL (0. 040 L) of 0. 100 M

barbituric acid (Ka = 9. 8  10–5) with 0. 100 M KOH.

(a) no KOH added

(b) 20 mL of KOH solution added

(c) 39 mL of KOH solution added

(d) 40 mL of KOH solution added

(e) 41 mL of KOH solution added

User AmitB
by
7.4k points

1 Answer

1 vote
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

User Juan Leni
by
7.4k points