Question

A 6.0 L waste solution in the lab contains F- at a concentration of 0.1045 M, but the disposal concentration limit for this anion is 1.3 mM. To reduce the concentration to the legal limit you decide to precipitate most of the fluoride by the addition of MgCl2 (molar mass=95.21 g/mol). A) what will be the equilibrium molar concentration of Mg2+ needed to reduce the [F-] concentration to 1.3 mM? B) what total mass of MgCl2 is needed to precipice the waste? (Ksp of MgF2 = 7.4 x 10^-9.)

Answer 1

The initial F^- concentration is 0.1045 M while the allowed F^- concentration is 1.3 mM. MgCl₂ is added to the waste to precipitate F^- as MgF₂ as per the reaction:

MgCl₂ + 2 F^- -----> MgF₂ + 2 Cl^-

As per the stoichiometric reaction above, 1 mole MgCl₂ = 2 moles F^- = 1 mole MgF₂.

1) MgF₂ is sparingly soluble and dissociates as

MgF₂ (s) <=====> Mg²⁺ (aq) + 2 F^- (aq)

K_sp = [Mg²⁺][F^-]²

Given [F^-] = 1.3 mM = 1.3*10^-3 M and K_sp = 7.4*10^-9, we have

7.4*10^-9 = [Mg²⁺]*(1.3*10^-3)²

===> 7.4*10^-9 = [Mg²⁺]*(1.69*10^-6)

===> [Mg²⁺] = (7.4*10^-9)/(1.69*10^-6) = 4.378*10^-3 ≈ 4.4*10^-3

The equilibrium molar concentration of Mg²⁺ is 4.4*10^-3 M = 4.4 mM (ans).

2) We desire to precipitate the entire F^- in the waste as MgF₂; calculate the moles of F^- precipitated = (volume of waste solution in L)*(concentration of F^-in mol/L) = (6.0 L)*(0.1045 mol/L) = 0.627 mol.

Use the stoichiometric reaction above.

0.627 mole F^- = (0.627 mole F^-)*(1 mole MgCl₂/2 moles F^-)*(95.21 g MgCl₂/1 moles MgCl₂) = 29.8483 g MgCl₂ (ans).