What is the minimum mass of Mg(NO₃)₂ that must be added to 1.00 L of a 0.260 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

HF is a weak monoprotic acid. From the concentration of HF and the pK_a, we need to calculate the concentration of F^-, which is needed for the precipitation question.

HF ⇔ H⁺ + F^-

Setup an ICE table to determine concentration of each species at equilibrium.

[HF] = 0.260 - x

[H⁺] = x

[F^-] = x

The K_a expression for HF dissociation is:

K_a = [H⁺][F^-]/[HF]

Assuming the value of x is negligible in the dissociation of HF at equilibrium, solve for [F^-] using the K_a expression.

7.2 x 10^-4 = x² / 0.260

x² = 1.872 x 10^-4

x = √1.872 x 10^-4 = 1.4 x 10^-2 M = [F^-]

In the 2nd reaction, Mg(NO₃)₂ reacts with HF to form insoluble MgF₂. The net ionic equation for this reaction is:

Mg²⁺ (aq) + 2F^- (aq) ⇔ MgF₂ (s)

K_sp = [Mg²⁺] [F^-]²

Solve for [Mg²⁺]

7.4 x 10^-9 = (x) (1.4 x 10^-2)²

x = 7.4 x 10^-9 / 1.96 x 10^-4 = 3.7 x 10^-5 M = [Mg²⁺]

Use the calculated molarity and volume (1.00 L) to determine moles of Mg²⁺.

3.7 x 10^-5 mol/L x 1.00 L = 3.7 x 10^-5 mol Mg²⁺

Molar mass of Mg(NO₃)₂ = 148 g/mol

mass of Mg(NO₃)₂ needed = 3.7 x 10^-5 mol x 148 g/mol = 0.0055 g Mg(NO₃)₂