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Tardigrade
Question
Chemistry
50 mL of 0.02 M NaOH solution is mixed 50 mL of 0.06 M acetic acid solution, the pH of resulting solution is ( p Ka. of acetic acid is 4.76, log 5=0.70 )
Q.
50
m
L
of
0.02
M
N
a
O
H
solution is mixed
50
m
L
of
0.06
M
acetic acid solution, the
p
H
of resulting solution is
(
p
K
a
of acetic acid is
4.76
,
lo
g
5
=
0.70
)
1923
195
AP EAMCET
AP EAMCET 2019
Report Error
A
5.06
33%
B
4.06
0%
C
5.46
67%
D
4.46
0%
Solution:
Given, Volume of
N
a
O
H
solution
=
50
m
L
Molarity of
N
a
O
H
solution
=
0.02
M
Meq. of
N
a
O
H
=
50
×
0.02
=
1
m
e
q
Volume of
C
H
3
COO
H
solution
=
50
m
L
Molarity of
C
H
3
COO
H
=
0.06
M
Meq. of
C
H
3
COO
H
=
50
×
0.06
=
3
m
e
q
1
meq. of
N
a
O
H
combines with
3
meq. of
C
H
3
COO
H
and forms
1
meq. of
C
H
3
COON
a
.
∴
2
meq. of
C
H
3
COO
H
is left.
So, it forms an acidic buffer.
Now, for an acidic buffer,
p
H
=
p
K
a
+
lo
g
[
Acid
Salt
]
p
H
=
p
K
a
+
lo
g
[
C
H
3
COO
H
]
[
C
H
3
COON
a
]
∴
p
H
=
4.76
+
lo
g
[
2
l
]
p
H
=
4.76
+
lo
g
(
5
×
1
0
−
1
)
p
H
=
4.76
+
lo
g
5
−
lo
g
10
=
4.76
+
0.70
−
1
=
4.46