Tardigrade
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Tardigrade
Question
Chemistry
Solubility of CaF2 is 0.5 × 10-4 mol L-1. The value of Ksp for the salt is
Q. Solubility of
C
a
F
2
is
0.5
×
1
0
−
4
m
o
l
L
−
1
. The value of
K
s
p
for the salt is
2524
200
Equilibrium
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A
5
×
1
0
−
12
11%
B
2.5
×
1
0
−
16
10%
C
1
×
1
0
−
13
5%
D
5
×
1
0
−
13
74%
Solution:
K
s
p
=
0.5
×
1
0
−
4
×
1
×
1
0
−
8
=
0.5
×
1
0
−
12
or
5
×
1
0
−
13