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Class 11 NEET Chemistry Equilibrium Questions Part 2
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© examsnet.com
Question : 19
Total: 24
pH of a saturated solution of
C
a
(
O
H
)
2
is 9. The solubility product
(
K
s
p
)
of
C
a
(
O
H
)
2
is:
[NEET 2019]
0.25
×
10
–
10
0.125
×
10
–
15
0.5
×
10
–
10
0.5
×
10
–
15
Validate
Solution:
👈: Video Solution
C
a
(
O
H
)
2
⇌
C
a
2
+
+
2
O
H
−
pH = 9, Hence
pOH = 14 – 9 = 5
[OH
–
] =
10
–
5
M
Hence
[
C
a
2
+
]
=
10
−
5
2
Thus
K
s
p
=
[
C
a
2
+
]
[
O
H
–
]
2
=
(
10
−
5
2
)
(
10
−
5
)
2
=
0.5
×
10
−
15
© examsnet.com
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