Question

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.

Answer 1

Answer:

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

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

Explanation:

• CaF2 ↔ Ca2+  +  2F-

          S             S          2S......in 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

• CaF2 ↔,Ca2+  +    2F-

         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