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Tardigrade
Question
Chemistry
pH of 0.0002 M formic acid [Ka=2× 10-4] approximately is
Q.
p
H
of
0.0002
M
formic acid
[
K
a
=
2
×
1
0
−
4
]
approximately is
3512
158
J & K CET
J & K CET 2012
Equilibrium
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A
1.35
B
0.5
C
3.7
D
1.85
Solution:
[
H
+
]
H
COO
H
=
K
a
.
C
=
2
×
1
0
−
4
×
0.0002
p
H
=
−
lo
g
[
H
+
]
=
−
lo
g
(
2
×
1
0
−
4
)
=
3.69
≈
3.7