Question

Ksp of srf2 in water is 10 8 10 . The solubility of srf2 in 0.1 m naf aqueous solution is

Answer 1

Answer:

Solubility of SrF₂ in 0.1 m NaF aqueous solution is = 8 X 10⁻⁸ mol/ L

Explanation:

Given

Ksp of SrF₂ in water = 8 X 10⁻¹⁰

Molarity of NaF = 0.1 M

Solubility of SrF₂ = ??

Solution

NaF dissociates into

NaF ---------> Na⁺ +  F⁻

0.1 M            0.1 M    0.1 M

similarly we can write

solubility (s)

SrF₂  -------------> Sr⁺² + 2F⁻

                          s           2s + 0.1 M

Therefore

Ksp = [Sr⁺² ] [F⁻]²

8 X 10⁻¹⁰ = s . (2s + 0.1)²

As 0.1 >>> 2s so 2s + 0.1 ≅ 0.1

8 X 10⁻¹⁰ = s (0.1)²

s = 8 X 10⁻¹⁰ X 0.01

s = 8 X 10⁻⁸ mol/ L

Answer 2

Answer:

see above attachment for solution