What is the minimum mass?

Question

What is the minimum mass of Mg(NO₃)₂ that must be added to 1.00 L of a 0.210 M HF solution to begin precipitation of MgF₂(s)? For MgF₂, Ksp = 7.4 × 10⁻⁹, and Ka for HF = 7.2 × 10⁻⁴.

Answer 1

First, let's find the [F-] present:

HF <==> H⁺ + F^-

Ka = [H⁺][F^-] / [HF]

7.2x10^-4 = (x)(x) / 0.210-x

x² = -7.2x10^-4x + 1.51x10^-4

x² + 7.2x10^-4x - 1.51x10^-4 = 0

x = [F^-] = 0.0119 M

MgF₂(s) ==> Mg²⁺(aq) + 2F^-(aq)

Ksp = 7.4x10^-9 = [Mg²⁺][F-]²

7.4x10^-9 = [Mg²⁺][0.0119]²

[Mg²⁺] = 5.23x10^-5 M

moles Mg²⁺ needed = 5.23x10^-5 moles since there is 1 liter of solution

mass Mg(NO₃)₂ = 5.23x10^-5 mols Mg²⁺ x 1 mol Mg²⁺ / mol Mg(NO₃)₂ x 148 g Mg(NO₃)₂ / mol =

7.74x10^-3 g Mg(NO₃)₂ = 7.74 mg as the minimum mass of Mg(NO₃)₂ that must be added

(be sure to check all of the math)