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Tardigrade
Question
Chemistry
The pH of 0.01 M KOH is
Q. The pH of
0.01
M
K
O
H
is
13038
178
COMEDK
COMEDK 2010
Equilibrium
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A
2
16%
B
8
15%
C
14
14%
D
12
55%
Solution:
pO
H
=
−
l
o
g
[
O
H
−
]
=
−
l
o
g
(
0.01
)
=
2
p
H
+
pO
H
=
14
p
H
=
14
−
pO
H
=
14
−
2
=
12