The equilibrium for the saturated solution of BaF^2 is 1.8*10^7. Calculate the solubility of BaF^2...

Answer 1

To find the solubility of BaF2, we need to set up an equilibrium expression using the solubility product constant, Ksp. The equation for the dissolution of BaF2 in water can be written as:

BaF2(s) ⇌ Ba²⁺(aq) + 2F⁻(aq)

Let's denote the solubility of BaF2 as x. This means that, at equilibrium, the concentration of Ba²⁺ ions is x and the concentration of F⁻ ions is 2x. Now we can write the Ksp expression:

Ksp = [Ba²⁺][sp = (x)(2x)²

We know that Ksp = 1.8 × 10⁻⁷, so we can now solve for x:

1.8 × 10⁻⁷ = (x)(2x)²
1.8 × 10⁻⁷ = 4x³

Now, we'll solve for x:

x³ = (1.8 × 10⁻⁷) / 4
x³ = 4.5 × 10⁻⁸

Take the cube root of both sides:

x = (4.5 × 10⁻⁸)^(1/3)
x ≈ 3.56 × 10⁻³

The solubility of BaF2 is approximately 3.56 × 10⁻³ M (mol/L).