Question

the solubility of BaF2 is 6.29x10^-3M and the solubility of barium fluoride in 0.15M NaF is 4.4x10^-5M. Compare the solubility and explain?

Answer 1

BaF2 ==> Ba^+2 + 2F^-

NaF ==> Na^+ + F^-

Solubility BaF2 alone = the above number
Solubility mixture BaF2 and 0.15 M NaF = a much smaller number. Why? Because of the common ion effect (the fluoride ion from NaF) shifts the BaF2 equilibrium to the left which results in a lower solubility of BaF2 in the mixture.