The ksp for calcium fluoride, caf2, is 1.5 x 10-10. a) excess solid calcium fluoride is added to 1.00 l of pure water. calculate the equilibrium concentration of calcium ion in the solution in moles/l. b) excess solid calcium fluoride is added to 1.00 l of 0.10 m sodium fluoride, naf. calculate the equilibrium concentration of calcium ion in the solution in moles/l.

a) [ Ca2+ ] = 3.347 E-4 mol/L

b) [ Ca2+ ] = 1.5 E-8 mol/L

Explanation:

          S             S          2Sin the equilibrium

⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S )² = 4S³

⇒ S = ∛ ( 1.5 E-10 / 4 )

⇒ S = ∛ 3.75 E-11

⇒ S = 3.347 E-4 mol/L

⇒ [ Ca2+ ] = S = 3.347 E-4 mol/L

b) NaF ↔ Na+  +  F-

  0.10 M    0.10     0.10

         S             S         2S + 0.10

⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S + 0.10 )²

∴∴ the Concentration: 0.10 M >>>> Ksp ( 1.5 E-10 ), son we can despise S as adding.

⇒ 1.5 E-10 = S * ( 0.10 )² = 0.01 S

⇒ S = 1.5 E-10 / 0-01

⇒ S = 1.5 E-8 mol/L

⇒ [ Ca2+ ] = S = 1.5 E-8 mol/L

explanation: