What is the minimum mass of Mg(NO₃)₂ that must be added to 1.00 L of a 0.760 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

Mg(NO₃)₂(aq) + 2HF(aq) ==> MgF₂(s) + 2HNO₃ ... balanced equation

First, find the [F-]:

HF ==> H⁺ + F-

Ka = 7.2x10^-4 = (x)(x) / 0.760 - x (assume x is small and ignore it in the denominator)

7.2x10-4 = x² / 0.760

x² = 5.47x10^-4

x = 0.0234 M = [F-]

MgF₂(s) <==> Mg²⁺(aq) + 2F-(aq)

Ksp = 7.4x10^-9 = [Mg²⁺][F-]²

7.4x10^-9 = [Mg²⁺][0.0234]² = [Mg²⁺][5.48x10^-4]

[Mg²⁺] = 1.35x10^-5 M

Converting this to moles / liter, we have...

molar mass Mg(NO₃)₂ = 148 g / mole

1.35x10^-5 mol / L x 148 g / mole = 2.00x10^-3 g = 2.00 mg of Mg(NO₃)₂