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Tardigrade
Question
Chemistry
100 mL of 0.02 M benzoic acid ( p Ka=4.2) is titrated using 0.02 M NaOH. pH after 50 mL and 100 mL of NaOH have been added are
Q.
100
m
L
of
0.02
M
benzoic acid
(
p
K
a
=
4.2
)
is titrated using
0.02
M
N
a
O
H
.
p
H
after
50
m
L
and
100
m
L
of
N
a
O
H
have been added are
1452
255
Equilibrium
Report Error
A
3.50,7
20%
B
4.2,7
20%
C
4.2,8.1
60%
D
4.2,8.25
0%
Solution:
When
50
m
L
of
0.02
M
N
a
O
H
is added inside, we will have [salt] = [acid] at half equivalence point,
p
H
=
p
K
a
+
lo
g
[
Acid
]
[
Salt
]
=
4.2
When
100
m
L
of
N
a
O
H
is added we will have equivalence point so
p
H
will be calculated according to hydrolysis of salt of weak acid and strong base.
p
H
=
2
1
(
p
K
w
+
p
K
a
+
lo
g
C
)
=
2
1
(
14
+
4.2
+
lo
g
0.02
)
=
8.25