Question

What is the molar solubility of    MgF2MgF2  in a 0.80 Mg(NO3)2Mg(NO3)2 solution? For MgF2,   Ksp = 8.4 ×10–8MgF2,   Ksp = 8.4 ×10–8.

Select one:

a.
8.0×10–8 M8.0×10–8 M

b.
2.3×10–4 M2.3×10–4 M

c.
2.0×10−8 M2.0×10−8 M

d.
1.6×10–4 M1.6×10–4 M

e.
3.2×10–3 M

Answer 1

MgF2     =     Mg²⁺     +     2F^-

Ksp = [Mg²⁺] [F^-]²

[F^-] = sqrt(Ksp / [Mg²⁺])

      = sqrt [ (8.4 x 10^-8) / 0.8) ]

     = sqrt (1.05 x 10^-7)

     = 3.24 x 10^-4

So, the solubility of MgF2 = [F^-] / 2

                                           = (3.24 x 10^-4) / 2

                                           = 1.62 x 10^-4

= 1.6 x 10^-4

Hence, the solubity of MgF₂ in 0.80 M Mg(NO3)2 = 1.62 x 10^-4 M

Answer is option (d).