Question

The Ksp for CaF2 is 3.9 x 10-11 at 25oC. What is the solubility?

Answer 1

Answer:

Option C.

Explanation:

To know this, we always has to write the chemical quation that is taking place:

CaF₂(s) <--------> Ca²⁺(aq) + 2F⁻(aq)    Ksp = 3.9x10⁻¹¹

The solubility of the solution, is given by the ions. This can be calculated using an ICE chart, and then, writting the equilibrium constant expression with these values. So, let's do the ICE chart:

        CaF₂(s) <--------> Ca²⁺(aq) + 2F⁻(aq)    Ksp = 3.9x10⁻¹¹

i)                                    0              0

c)                                  +s            +2s

e)                                   s               2s

Now, let's write the Ksp expression using these values:

Ksp = [Ca²⁺] [F⁻]²

Replacing the values, and solving for solubility "s" we have:

3.9x10⁻¹¹ = s * (2s)²

3.9x10⁻¹¹ = s * 4s²

3.9x10⁻¹¹ = 4s³

s³ = 3.9x10⁻¹¹ / 4

s = ∛(3.9x10⁻¹¹/4)

s = 2.14x10⁻⁴ M

This is the solubility of the solution. Option C.

Answer 2

We have to know the solubility of CaF₂.

The solubility of CaF₂ is: (c) 2.1 x 10-4 Molar

The general expression of solubility product of any sparingly soluble salt (having solubility S) with formula $A_{x} B_{y}$ is: $x^{x} y^{y} S^{x+y}$.

For the compound, CaF₂, x=1, y=2 So, $K_{sp}$=$1^{1}2^{2}S^{1+2}$=4S³= 3.9 x 10-11  (Given)

S³=$\frac{3.9 X 10^{-11} }{4}$

S³=9.75  X $10^{-12}$

S= 2.1 X $10^{-4}$ Molar