Ionic EquilibriumHard
Question
The solubility of CaF2 (Ksp = 3.4 × 10-11) in 0.1 M solution of NaF would be
Options
A.3.4 × 10-12 M
B.3.4 × 10-10 M
C.3.4 × 10-9 M
D.3.4 × 10-13 M
Solution
(a) NaF Na+ + F-
0.1 0.1 0.1
CaF2 Ca2+ + 2F-
(2x + 0.1) ≈ 0.1
Ksp = x (0.1)2 = 3.4 × 10-11 x = 3.4 × 10-9
0.1 0.1 0.1
CaF2 Ca2+ + 2F-
(2x + 0.1) ≈ 0.1
Ksp = x (0.1)2 = 3.4 × 10-11 x = 3.4 × 10-9
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