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The solubility of CaF2 is 0.00021 mole per liter. What is the solubility product constant for CaF2? A) 7.3 x 10-12 B) 3.7 x 10-11 C) 8.5 x 10-8 D) 4.4 x 10-8 E) 1.9 x 10-11

Respuesta :

Answer:

B) Ksp = 3.7 E-11

Explanation:

CaF2 ↔ Ca2+  +  2F-

  S            S          2S,,,,,,,,,,,,,,,in the equilibrium

⇒ Ksp = [ Ca2+ ] * [ F- ]²

⇒ Ksp = S * ( 2S )²

⇒ Ksp = 4S³

⇒ Ksp = 4 * ( 0.00021 )³

⇒ Ksp = 4 * 9.261 E-12

⇒ Ksp = 3.704 E-11

Eduard22sly

The solubility product constant for CaF₂ is 3.7×10¯¹¹.

The correct answer to the question is Option B. 3.7×10¯¹¹

We'll begin by writing the balanced dissociation equation for CaF₂. This is illustrated below:

CaF₂ <=> Ca²⁺ + 2F¯

At equilibrium,

0.00021 mol/L of CaF₂ will produce:

  • 0.00021 mol/L of Ca²⁺
  • 2 × 0.00021 = 0.00042 mol/L of F¯

With the above information, we can obtain the solubility product constant for CaF₂. This can be obtained as follow:

Concentration of calcium ion [Ca²⁺] = 0.00021 mol/L

Concentration of fluoride ion [F¯] = 0.00042 mol/L

Solubility product constant (Ksp) =?

CaF₂ <=> Ca²⁺ + 2F¯

Ksp = [Ca²⁺] [F¯]²

Ksp = 0.00021 × (0.00042)²

Ksp = 3.7×10¯¹¹

Therefore, the solubility product constant for CaF₂ is 3.7×10¯¹¹.

The correct answer to the question is Option B. 3.7×10¯¹¹

Learn more: https://brainly.com/question/1385943

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