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Ionic Equilibrium I
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Solubility product and simple solubility calculations
JEE Main level
Single correct option
Difficulty 2 of 3
ie184
The solubility of
PbCl
X
2
(molar mass
278
g
mo
l
−
1
) in water is
4.17
g
L
−
1
. Its
K
s
p
is approximately
A
2.2
×
1
0
−
4
B
3.4
×
1
0
−
6
C
7.3
×
1
0
1
D
1.3
×
1
0
−
5
Answer: D
s
=
4.17/278
=
1.5
×
1
0
−
2
M
. For
PbCl
X
2
,
K
s
p
=
s
(
2
s
)
2
=
4
s
3
=
1.3
×
1
0
−
5
.
Practice Ionic Equilibrium I
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