Molar solubility!! chemistry?

Let s M be the molar solubility of CaF₂.

Initial concentration of Ca²⁺, [Ca²⁺]ₒ = [Ca(NO₃)₂]ₒ = 0.25 M

Consider the solubility equilibrium of CaF₂ :
_________ CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq) ___ Ksp = 1.46 × 10⁻¹⁰
Initial: ____________ 0.25 M ___ 0 M
Change: ___________ +s M ___ +2s M
At eqm: ________(0.25 + s) M __ 2s M

As Ksp is very small and due to the common ion effect in the presence of Ca²⁺ ions, the solubility of CaF₂ is very small.
It can be assumed that 0.25 ≫ s, and thus [Ca²⁺] at equilibrium = (0.25 + s) M ≈ 0.25 M

At equilibrium :
Ksp = [Ca²⁺] [F⁻]²
1.46 × 10⁻¹⁰ = 0.25 × (2s)²
1.46 × 10⁻¹⁰ = s²
s = √(1.46 × 10⁻¹⁰) = 1.2 × 10⁻⁵

Solubility of CaF₂ in 0.25 M Ca(NO₃)₂ = 1.2 × 10⁻⁵ M