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Q.

The solubility of a saturated solution of calcium fluoride is 2x10-4 mol L-1. Its solubility product is

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a

12×10−2

b

14×10−4

c

22×10−11

d

32×10−12

answer is D.

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Detailed Solution

CaF2⇌Ca2+(aq) +2F−(aq)                     S               2S Ksp=Ca2+F−2=(S)(2.S)2=4.S3 =42×10−43=32×10−12M3
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