The solubility of CaF2 (Ksp = 3.4 × 10-11) in 0.1 M solution of NaF would be

(a) NaF Na+ + F- 0.1 0.1 0.1 CaF2 Ca2+ + 2F- (2x + 0.1) ≈ 0.1Ksp = x (0.1)2 = 3.4 × 10-11 x = 3.4 × 10-9

Ionic EquilibriumHard

Question

The solubility of CaF₂ (K_sp = 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
K_sp = x (0.1)² = 3.4 × 10^-11    x = 3.4 × 10^-9

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