Does a precipitate of magnesium fluoride form if 300.0 mL of 1.1 × 10^-3 M MgCl₂ are added to 500.0 mL of 1.2 × 10^-3 M NaF?
[K_sp (MgF₂) = 6.9 × 10^-9]

Question

Does a precipitate of magnesium fluoride form if 300.0 mL of 1.1 × 10^-3 M MgCl₂ are added to 500.0 mL of 1.2 × 10^-3 M NaF? [K_sp (MgF₂)= 6.9 × 10^-9]

Quizplus · Accepted Answer

To determine if a precipitate of magnesium fluoride will form, we need to calculate the ion product of magnesium fluoride, Q, and compare it to the solubility product constant, Ksp.

The balanced chemical equation for the reaction is:

Mg2+(aq) + 2F-(aq) → MgF2(s)

The initial concentrations of Mg2+ and F- ions are:

[Mg2+] = 1.1 × 10^-3 M
[F-] = 1.2 × 10^-3 M

Using the stoichiometry of the balanced equation, we can see that the concentration of Mg2+ ions is in excess:

[Mg2+]excess = [Mg2+] - 0 = 1.1 × 10^-3 M

Therefore, the maximum amount of MgF2 that could possibly form is limited by the concentration of F- ions remaining. Assume that x is the molar solubility of MgF2 in water. Then:

[Mg2+] = x M
[F-] = 1.2 × 10^-3 M - 2x M

The ion product Q is:

Q = [Mg2+][F-]^2 = x(1.2 × 10^-3 - 2x)^2

Substituting the values given, we get:

Q = (1.1 × 10^-3)(1.2 × 10^-3)^2 = 1.584 × 10^-9

Comparing Q to the solubility product constant Ksp:

Ksp = 6.9 × 10^-9

We see that Q < Ksp. This means that MgF2 will not precipitate out of solution since there is still room for more MgF2 to dissolve. Therefore, the answer is B.

Does a Precipitate of Magnesium Fluoride Form If 300