Question

Calculate the molar solubility of barium fluoride in 0.15 M Nag, Ksp for BaF2 is 2.45×10^-5

Answer 1

There is a typo in the question and it should be as follows:

Calculate the molar solubility of barium fluoride in 0.15 M NaF, Ksp for BaF2 is 2.45×10^-5.

Solubility product constant (Ksp)  indicates the degree to which a compound dissociates in water. It is simplified equilibrium constant defined for equilibrium between a solid and its respective ions in a solution. Highly soluble compounds will have an higher value for the solubility product constant (Ksp).

For the given problem the dissociation of BaF2 is given as follows:

BaF2(s) <----------------------> Ba²⁺​ (aq) + 2F^​- (aq)

The Ksp expression for a salt is the product of the concentrations of the ions, with each concentration raised to a power equal to the coefficient of that ion in the balanced equation for the solubility equilibrium.

Ksp = [Ba²⁺​ ​] [ F^​-​ ]^​2

Note that the concentration of solid [BaF2] itself does not appear as a denominator in the expression since it is not in the same phase as the aqueous ions unlike in equilibrium constant expression.

In 0.15 M NaF solution Ksp for BaF2 is 2.45×10^-5. So in the above expression we need to take into consideration of Fluoride ion concentration [F^-​] from dissolution of NaF also. Accordingly if x is the concentration of Ba²⁺​ ions then 2x + 0.15 will be the concentration of F^-​ ions.

BaF2(s) <----------------------> Ba²⁺​ (aq) + 2F^​- (aq)

                                                 x                2x + 0.15

Since  BaF2 is sparingly soluble in water, we might expect "2x" to be not significant compared to 0.15. In that case one may ignore 2x in 2x + 0.15 which then equals to 0.15 alone and substituting these values into Ksp expression we get following:

         Ksp = [Ba²⁺​ ​] [ F^​-​ ]^​2
2.45×10^-5​ = x [0.15]² and solving for x we get

              x = 2.45×10^-5​ / [0.15]²

​                 = 2.45 x 10^-5​  / 0.0225

                 = 108.88 x 10^-5​

             x = 1.08 x 10^-3​ M.

Hence molar solubility of Barium Fluoride in 0.15M NaF is 1.08 x 10^-3​ M. As can be seen 2x is much smaller than 0.15 and our assumption is correct in this regard.