Question

Calculate the molar solubility of magnesium fluoride in each of the following solutions: (MgF2 Ksp = 7.4*10^-9)
a) pure water
b) 0.010 M potassium fluoride solution
c) 0.050 M solution of magnesium nitrate
d) Why are the results of parts (a) thru (c) different?

Answer 1

Answer – Given, Ksp of magnesium fluoride = 7.4*10^-9

Molar solubility = ?

a)Pure water

In the pure water there is only Mg²⁺ and 2F^-

So Kap = [Mg²⁺] [F^-]²

7.4*10^-9 = x* (2x)²

4x³ = 7.4*10^-9

x³ = 7.4*10^-9 /4

       = 1.85*10^-9

So, x = 0.00123 M

So molar solubility of magnesium fluoride is 0.00123 M in pure water .

b) 0.010 M potassium fluoride solution

[KF] = [F^-] = 0.010

So, Kap = [Mg²⁺] [0.010]²

7.4*10^-9 = x* 0.0001

x = 7.4*10^-9 /0.0001

x³ = 7.4*10^-9 /4

       = 7.4*10^-5

Molar solubility of magnesium fluoride is 7.4*10^-5 M in 0.010 M potassium fluoride solution.

C) 0.050 M solution of magnesium nitrate

[Mg(NO₃)₂] = [Mg²⁺] = 0.050

So, Kap = [Mg²⁺] [F^-]²

7.4*10^-9 = x* (2x)²

4x² = 7.4*10^-9 /0.05

x³ = 1.48*10^-8 /4

       = 3.7*10^-8

x = 1.92*10^-4 M

Molar solubility of magnesium fluoride is 1.92*10^-4 M in 0.050 M solution of magnesium nitrate

d) The result for the a is in the pure water, so there is no any common ion, but in the b and c there are added solution that as common ion effect and that decrease the molar solubility of MgF₂, since there is more reactant added there is formed more product and reverse reaction. So that’s why there is difference in the result.