The equilibrium for the saturated solution of BaF^2 is 1.8 *10^-7
Calculate the solubility of BaF^2 in moles per liter
To calculate the solubility of BaF2, we can use the equilibrium expression for its dissolution:
BaF2(s) ⇌ Ba²⁺(aq) + 2F⁻(aq)
The equilibrium constant expression for this reaction is:
Ksp = [Ba²⁺][F⁻]^2
We are given Ksp = 1.8 × 10⁻⁷. Let the solubility of BaF2 in mol/L be S. Since the stoichiometry of the reaction indicates that 1 mole of BaF2 produces 1 mole of Ba²⁺ and 2 moles of F⁻, we can write the concentrations of Ba²⁺ and F⁻ in terms of S:
[Ba²⁺] = S
[F⁻] = 2S
Now we can substitute these expressions into the Ksp expression:
Ksp = (S)(2S)^2
1.8 × 10⁻⁷ = S(4S^2)
1.8 × 10⁻⁷ = 4S^3
Divide both sides by 4:
4.5 × 10⁻⁸ = S^3
Now take the cube root of both sides to solve for S:
S = (4.5 × 10⁻⁸)^(1/3)
S ≈ 1.65 × 10⁻³
The solubility of BaF2 is approximately 1.65 × 10⁻³ mol/L.