Final answer:
The molar solubility of magnesium fluoride in a solution that is 0.40 M F- ions is approximately 7.09 × 10-7 M.
Step-by-step explanation:
In order to determine the molar solubility of magnesium fluoride (MgF2) in a solution containing 0.40 M F- ions, we can use the solubility product constant (Ksp) of MgF2 and the stoichiometry of the dissolution equation. The solubility product constant (Ksp) is a mathematical constant that represents the equilibrium between the dissolved ions of a sparingly soluble salt in a saturated solution. For MgF2, the Ksp value is given as 6.4 × 10-9.
To find the molar solubility, we need to consider the equilibrium equation representing the dissolution of MgF2:
MgF2 (s) → Mg2+ (aq) + 2F- (aq)
The stoichiometry of the dissolution equation tells us that the concentration of fluoride ions (F-) is twice the concentration of magnesium ions (Mg2+). Therefore, if we let x represent the molar solubility of MgF2, the concentration of F- ions would be 2x.
Using the Ksp expression, we can set up the equation:
Ksp = [Mg2+][F-]2 = (x)(2x)2
Substituting the given value of Ksp into the equation:
6.4 × 10-9 = (x)(4x2)
Simplifying the equation:
6.4 × 10-9 = 4x3
Since we are looking for the molar solubility, which is the concentration of MgF2 in moles per liter, we will assume that x is small and can be neglected in comparison to the initial concentration of F- ions (0.40 M). Therefore, we can use the approximation:
x << 0.40 M
By substituting this approximation into the equation, we can solve for x:
6.4 × 10-9 = 4x3
Solving for x:
x ≈ 7.09 × 10-7 M
Therefore, the molar solubility of MgF2 in a solution containing 0.40 M F- ions is approximately 7.09 × 10-7 M.