Question

Calculate the molar solubility of magnesium fluoride in each of the following solutions:
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 – In this question we are given different molarity of solution. We know the Ksp of magnesium fluoride and it is 3.7*10^-8

a)Pure water – We know in the pure water there are Mg²⁺ and 2F^-

we know the Ksp expression for MgF₂

Kap = [Mg²⁺] [F^-]²

3.7*10^-8 = x* (2x)²

4x³ = 3.7*10^-8

x³ = 3.7*10^-8 /4

       = 9.25*10^-9

x = 0.00210 M

The molar solubility of magnesium fluoride is 0.00210 M in pure water .

b) Molar solubility of magnesium fluoride in 0.010 M potassium fluoride solution

[KF] = [F^-] = 0.010

Kap = [Mg²⁺] [0.010]²

3.7*10^-8 = x* 0.0001

x = 3.7*10^-8 /0.0001

x = 3.7*10^-4

Molar solubility of magnesium fluoride is 3.7*10^-4M in 0.010 M potassium fluoride solution.

C) ) Molar solubility of magnesium fluoride in 0.050 M solution of magnesium nitrate

[Mg(NO₃)₂] = [Mg²⁺] = 0.050

Kap = [Mg²⁺] [F^-]²

3.7*10^-8 = x* (2x)²

3.7*10^-8 = 0.05* (2x)²

4x² = 3.7*10^-8 /0.05

x² = 7.4*10^-7 /4

       = 1.85*10^-7

x = 0.00043 M

Molar solubility of magnesium fluoride is 0.00043 M in 0.050 M solution of magnesium nitrate

d) In the b and c there are showing as common ion effect and that decrease the molar solubility of MgF₂, but in the result for the a is in the pure water, so there is no any common ion. We know when there is more reactant added there is formed more product and reverse reaction and molar solubility gets decreased, so as the we are getting the different result for a to c.