Question

A saturated solution of magnesium fluoride , mgf2, was prepared by dissolving solid mgf2 in water. the concentration of mg2+ ion in the solution was found to be 1.18×10−3 m . calculate ksp for mgf2.

Answer 1

Ksp = [Mg⁺²] [2F⁻]²
       = (1.18 x 10⁻³) (2 x 1.18 x 10⁻³)² 
       = 6.57 x 10⁻⁹ 

Answer 2

Answer: Ksp = 6.57 × 10 ⁻⁹

Explanation:

1) Ksp is the solubility product constant.

2) Ksp is related with the equilibrium chemical equation in this way:

i) equilibrium chemical equation: cC + dD ⇄ aA + bB

ii) Ksp = [A]ᵃ [B]ᵇ ← here A and B are the products not the reactants

Note: in Ksp problems, the reactants do not form part of the Ksp equation, because the solubilities are so small that the concentrations of the reactants are condiered constant.

3) The dissolving equaton for MgF₂

MgF₂ ⇄ Mg²⁺ + 2F⁻

From it you can calculate the species in equilibrium:

     A - x                ⇄         x           +              x

A - 0.00118                0.00118                 2 × 0.00118

x = 0.00118

4) Therefore the Ksp equation is:

Ksp = [Mg²⁺] [F⁻}² = (x)(2x²) = 4x³

Ksp = 4(0.00118)² = 6.57 × 10 ⁻⁹