Calculate the molar solubility of CaF2 in a 0.2890 M cesium fluoride, CsF solution.?

Let s M be the molar solubility of CaF₂ in a 0.2890 M CsF solution.

CsF(s) ___ ⇌ ___ Cs⁺(aq) ___ + ___ F⁻(aq) ____ Ksp = 3.50 × 10⁻¹¹
Initial: _________ 0 M _________0.2890 M
Change: _______ +s M ___________+s M
Equilibrium: ____ s M _______ (0.2890 + s) M

As Ksp of CaF₂ is very small and due to the common ion effect in the presence of F⁻,
the molar solubility of CaF₂ is very small.
It can be assumed that 0.2890 ≫ s and thus [F⁻] at equilibrium = (0.2890 - s) M ≈ 0.2890 M

At equilibrium:
Ksp = [Cs⁺] [F⁻]
3.50 × 10⁻¹¹ = s × 0.2890
s = (3.50 × 10⁻¹¹) / 0.2890 = 1.21 × 10⁻¹⁰

Molar solubility of CaF₂ in a 0.2890 M CsF solution = 1.21 × 10⁻¹⁰ M

CaF2(s) ⇐=> Ca2+ + 2F-

Ksp = [Ca2+][F-]^2 = 3.50 x 10^-11

Let S = molar solubility of CaF2

CsF ==⇒ Cs+ + F-
Initial [F-] from CsF is 0.289

[Ca2+] = S ; [F-] = 0.289 + 2S

S(0.289 + 2S)^2 = 3.50 x 10^-11
Since the F- from the CsF is so much greater than the 2S
from the CaF2 and the Ksp value is very small, one can neglect the 2S.

S(0.289)^2 = 3.50 x 10^-11
S = 4.19 x 10^-10 which is the molar solubility

No