Question

1. Magnesium (II) fluoride, MgF2, is a sparingly soluble salt in water. The value of the solubility product constant, Ksp, is 6.4⨉10^-9 at 298 K

(a) Write the equilibrium reaction for the dissolution reaction of MgF2 in water.

(b) Write an expression for the solubility product, Ksp, in terms of the ion concentrations

(c) Calculate the molar solubility of MgF2 in water at 298 K.

(d) 20 cm3 of a solution containing Mg2+(aq) ions at a concentration of 0.02 mol dm-3 is added to 10 cm3 of a solution containing F-(aq) ions at a concentration of 0.02 mol dm-3. Predict, with reasoning, whether a MgF2 precipitate will form.

Answer 1

a) MgF2(s) ------------------ Mg+2(aq) + 2 F-(aq)

b) Ksp = [Mg+2][F-]^2

c)      MgF2(s) ------------------ Mg+2(aq) + 2 F-(aq)

                                                S                   2S               where S= solubility

         Ksp = S x(2S)^2

          Ksp = 4S^3

          S^3 = Ksp/4 = 6.4x10^-9/4 = 1.6x10^-9

        S = 1.169 x10^-3

Molar solubility of MgF2 = S= 1.17 x10^-3 M

d)

Mg+2 = 20 cm3 of 0.02 mol/dm-3

number of moles of Mg+2 ions = 0.02 mol/dm-3 x 0.020 dm-3

number o fmoles of Mg+2 = 0.0004 mole

F- = 10 cm-3 of 0.02 mol dm-3

number of moles of F- = 0.02 moldm-3 x0.010 dm-3

number of moles of F- = 0.0002 mol

MgF2(s) ------------------ Mg+2(aq) + 2 F-(aq)

Q= [Mg+2][F-]^2

Q= ( 0.0004)x(2x0.0002)^2

Q= 6.4x10^-11

Ksp of MgF2 = 6.4x10^-9

Q < Ksp

No, Precipitate will form.