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Question
Chemistry
The pOH of 0.0005 M sulphuric acid is
Q. The pOH of
0.0005
M
sulphuric acid is
2804
221
COMEDK
COMEDK 2011
Equilibrium
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A
5
16%
B
3
36%
C
11
34%
D
12
14%
Solution:
H
2
S
O
4
⇌
2
H
+
+
S
O
4
2
−
So,
[
H
+
]
=
2
×
0
⋅
0005
=
1
×
1
0
−
3
M
p
H
=
−
lo
g
[
H
+
]
=
−
lo
g
(
1
×
1
0
−
3
)
p
H
=
3
∵
p
H
+
pO
H
=
14
3
+
pO
H
=
14
pO
H
=
14
−
3
pO
H
=
14
pO
H
=
11