Question bank/JEE Main/Chemistry

The solubility of CaF2 (Ksp = 5.3 × 10-9) in 0.1 M solution of
NaF would be: Assume no reaction of cation/anion. n × 10-7. Find the
value of n.
NaF ds 0.1 M CaF2 (Ksp = 5.3 × 10-9).......
gksxh u ekusaA n × 10-7. n

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(a) NaF[IMAGE] Na+ + F-
0.1 0.1 0.1
CaF2 [IMAGE] Ca2+ + 2F-
x (2x + 0.1) 0.1
Ksp = x (0.1)2 = 5.3 × 10-9
So vr, x = 5.3 × 10-7 M.

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx