Question

The solubility of CaF2 In a solotion of 0.1M of Ca(No3)2 is given by

Answer 1

Answer:

The solubility of CaF2 (s) [Ksp= 4.0 x 10 -11] in 1.0 L of a 1.0 x 10 -2 M solution of NaF is: A. 4.0 x 10 -9 moles B. 1.0 x 10 -7 moles C.Write rxn: CaF2(s) ⇌ Ca2+(aq) + 2 F–(aq) 2. Write Ksp: Ksp = [Ca2+][F–]2 = 3.9 × 10–11 3. ICE Table 4. Substitute the equilibrium concentration values from the table into the.

Answer 2

The solubility of CaF₂ in a solution of 0.1 M is 3.08 × 10⁻⁶ M.

Explanation:

The chemical equation of the reaction is given as:

CaF₂ ⇄ Ca⁺² + 2F⁻

When 1 mole of salt is dissolved then 1 mole of Ca⁺² and 2 moles of F⁻ is formed.

The Ksp is given as:

Ksp = [Ca⁺²] × [F⁻]²

Where,

[Ca⁺²] = (0.1 + S) M

[F⁻] = S

Ksp = 3.9 × 10⁻¹¹

On substituting the values, we get,

3.9 × 10⁻¹¹ = (0.1 + S) + 4S²

3.9 × 10⁻¹¹ = 0.1 + 4S²

3.8 × 10⁻¹¹ = 4S²

S² = (3.8 × 10⁻¹¹)/4

∴ S = 3.08 × 10⁻⁶ M