Tardigrade
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Tardigrade
Question
Chemistry
pH of 0.005 M calcium acetate ( p Ka. of CH 3 COOH =4.74 ) is :
Q.
p
H
of
0.005
M
calcium acetate
(
p
K
a
of
C
H
3
COO
H
=
4.74
) is :
4882
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A
7.04
B
9.37
C
9.26
D
8.37
Solution:
0.005
M
calcium acetate
(
C
H
3
COO
)
2
C
a
0.005
M
(
C
H
3
COO
)
2
C
a
→
C
a
2
+
+
2
×
0.005
=
0.01
)
2
C
H
3
CO
O
−
∴
[
C
H
3
CO
O
−
]
=
0.01
M
C
H
3
CO
O
−
+
H
2
O
C
H
3
COO
H
+
Alkaline
O
H
−
p
H
=
7
+
2
p
K
a
+
2
l
o
g
C
=
7
+
2.37
+
2
l
o
g
0.01
=
7
+
2.37
−
1
=
8.37