Tardigrade
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Tardigrade
Question
Chemistry
0.001 mol of [ Co ( NH 3)5( NO 3)( SO 4)] was passed through a cation exchanger and the acid coming out of it required 20 mL of 0.1 M NaOH for neutralisation. Hence, the complex is
Q.
0.001
m
o
l
of
[
C
o
(
N
H
3
)
5
(
N
O
3
)
(
S
O
4
)
]
was passed through a cation exchanger and the acid coming out of it required
20
m
L
of
0.1
M
N
a
O
H
for neutralisation. Hence, the complex is
2219
222
Coordination Compounds
Report Error
A
[
C
o
(
N
H
3
)
5
(
S
O
4
)
]
N
O
3
B
[
C
o
(
N
H
3
)
5
(
N
O
3
)
]
S
O
4
C
[
C
o
(
N
H
3
)
5
]
N
O
3
⋅
S
O
4
D
none of these.
Solution:
Millimoles of acid
=
20
×
0.1
=
2
millimoles
=
0.002
moles
If
0.001
m
o
l
is neutralising
0.002
moles of base. Thus, acid coming out should be dibasic.
[
C
o
(
N
H
3
)
5
(
N
O
3
)
]
S
O
4
→
[
C
o
(
N
H
3
)
5
(
N
O
3
]
2
+
+
S
O
4
2
−