Question

The molar solubility of BaF₂ is 7.5 x 10^-3 mol/L. What is the value of Ksp for BaF₂?

Answer 1

Ans. Given, molar solubility of BaF2 = 7.5 x 10^-3 mol/ L = 7.5 x 10^-3 M. [Note that, 1 molar or 1 M = 1 mol/L]. It means that there is 7.5 x 10^-3 mol of BaF₂ solvated per liter of saturated solution.

Step 1. Write a balanced reaction for solvation of solid BaF2 into aqueous form. Note that the forward reaction shall favor solvation as follow-

            BaF2(s) + H2O --------------> Ba²⁺(aq ) + 2 F^-(aq)

Step 2: Calculate concentrations of solvation ions.

Stoichiometry: See the reaction above, 1 ml BaF₂ produces 1 mol Ba²⁺ and 2 mol F^- ions when solvated in water.

            So, [Ba²⁺] = [BaF₂] = 7.5 x 10^‑3 M

                        [F^-] = 2 x [BaF₂] = 2 x (7.5 x 10^‑3 M) = 1.5 x 10^-2 M

Step 3: Calculating Ksp

Ksp expression is equivalent to reaction quotient when a solid form is being converted into aqueous form.

            Ksp = [Ba²⁺] [F^-]² / [BaF₂] [H₂O]

Note that activity of solid reactant (BaF₂) and pure solvent (H₂O) is taken to be unity (= 1). So, those two terms at denominator are equal to 1, thus not written in any Ksp expression. Or, the Ksp expression consists of only multiplication of resultant solvated ions raised to the power of their respective moles.

So,

Ksp = [Ba²⁺] [F^-]²    

            Or, Ksp = [7.5 x 10^-3] [1.50 x 10^-2]² = 1.69 x 10^-9

Therefore, Ksp = 1.69 x 10^-6