Calcium fluoride (caf2) has a solubility constant of 3.45 x 10-11
. what is the molar solubility of caf2 in water?

4.15 x 10-6 m
5.87 x 10-6 m
2.05 x 10-4 m
3.26 x 10-4 m

Ksp=3.45×10⁻¹¹

CaF₂(s) ⇄ Ca²⁺(aq) + 2F⁻(aq)

Ksp=[Ca²⁺][F⁻]²

[Ca²⁺]=C(CaF₂)
[F⁻]=2C(CaF₂)

Ksp=4{C(CaF₂)}³

C(CaF₂)=∛(Ksp/4)

C(CaF₂)=∛(3.45×10⁻¹¹/4)=2.05×10⁻⁴ mol/L

2.05×10⁻⁴ M

Answerδh > 0explanationc₆h₆  (s)  → c₆h₆ (l)if you are changing it from a solid to a liquid, you are pretty much melting the solid. this requires heat energy. therefore, the reaction is endothermic (takes in energy); the system enthalpy increases as the reaction proceeds: δh > 0.