Question

What is the solubility, in milligrams per milliliter, of BaF2 in water containing 5.5 mg/mL KF (Ksp = 1.7 × 10-6)?

Answer 1

concventration of KF in water = 5.5/58 = 0.095 M

   concventration of F- in solution = 0.095 M

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

[Ba^2+] = x , [F-] = 0.095 M

Ksp = [Ba^2+][F-]^2

(1.7 *10^-6) = x*(0.095)^2

x = molarsolubility of BaF2 = 1.88*10^-4 M

solubility of BaF2 in mg/ml = 1.88*10^-4 *Mwt

                             = 1.88*10^-4 *175.3

                             = 0.033 mg/ml