Question

For the slightly soluble ionic compound MgF2, write the equilibrium equation for its dissociation:

a) MgF2(s) ⇌ Mg2+(aq) + 2F-(aq)
b) MgF2(s) ⇌ Mg2+(aq) + 2F2-(aq)
c) MgF2(s) ⇌ Mg2+(aq) + 2F-(aq) + 2e-
d) MgF2(s) ⇌ Mg2+(aq) + F-(aq) + F2-(aq)

Answer 1

The equilibrium equation for the dissociation of MgF2 is a) MgF2(s) ⇌ Mg2+(aq) + 2F-(aq), and the expression for the solubility-product constant, Ksp, can be determined and calculated using known concentrations of the ions in solution.

Step-by-step explanation:

The correct equilibrium equation for the dissociation of the slightly soluble ionic compound MgF2 is a) MgF2(s) ⇌ Mg2+(aq) + 2F-(aq).

AP Chemistry Problem Solving

Calculating Solubility-Product Constant, Ksp

Ksp can be calculated using the formula:
Ksp = [Mg2+][F-]^2
At 18 °C, with the concentration of Mg2+ being 1.21 × 10⁻³ M, the concentration of F- would be twice that since for every 1 mole of MgF2 that dissolves, 2 moles of F- ions are produced. Therefore, Ksp = (1.21 × 10⁻³) × (2 × 1.21 × 10⁻³)^2.