Question

Will a precipitate form when a 74.8mL sample of 0.0025 M KF is added to a beaker containing 45.5mL sample of 0.00148 M BaCl₂?

Answer 1

concentration of [F-] = 74.8 x 0.0025 / (74.8 + 45.5)

                                 = 1.55 x 10^-3 M

concentration of [Ba+2] = 45.5 x 0.00148 / (45.5 + 74.8)

                                      = 5.6 x 10^-4 M

BaF2 --------------> Ba+2 + 2 F-

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

       = (5.6 x 10^-4)(1.55 x 10^-3)^2

Kip = 1.34 x 10^-9   (ionic product)

Ksp of BaF2 = 1.84 x 10^-7 (solubility product)

here Ksp > Kip so

no precipitate will form