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Tardigrade
Question
Chemistry
20 mL of 0.1 M acetic acid is taken in a beaker. To this, 10 mL of 0.1 M NaOH solution is mixed. The pH of the resulting solution is. [ pK a . of acetic acid .=4.74]
Q.
20
m
L
of
0.1
M
acetic acid is taken in a beaker. To this,
10
m
L
of
0.1
M
N
a
O
H
solution is mixed. The
p
H
of the resulting solution is______.
[
p
K
a
of acetic acid
=
4.74
]
361
163
Equilibrium
Report Error
Answer:
4.74
Solution:
Moles of acetic acid
=
0.1
×
20/1000
=
0.002
m
o
l
Moles of
N
a
O
H
=
0.1
×
10/1000
=
0.001
m
o
l
p
H
=
p
K
a
+
lo
g
10
[
acid
]
[
salt
]
(Henderson's equation)
∴
p
H
=
4.74
+
lo
g
10
(
0.001
)
(
0.001
)
=
4.74
+
lo
g
10
1
=
4.74
+
0
=
4.74
