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JEE Main Chemistry Class 11 Equilibrium Questions Part 3 Questions
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© examsnet.com
Question : 29
Total: 33
25.0
mL
of
0.050
MBa
(
NO
3
)
2
is mixed with
25.0
mL
of
0.020
M
NaF
.
K
s
p
of
BaF
2
is
0.5
×
10
−
6
at
298
K
. The ratio of
[
Ba
2
+
]
[
F
−
]
2
and
K
s
p
is _________ . (Nearest integer)
[13-Apr-2023 shift 1]
Your Answer:
Validate
Solution:
[
Ba
+
2
]
=
25
×
0.05
50
=
0.025
M
[
F
−
]
=
25
×
0.02
50
=
0.01
M
[
Ba
+
2
]
[
F
−
]
2
=
25
×
10
−
7
K
sp
=
5
×
10
−
7
(given)
Ratio
=
[
Ba
+
2
]
[
F
−
]
2
K
s
p
=
5
© examsnet.com
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