Explanation:
BaF2(s) <------> Ba2+(aq) + 2F-(aq)
Ksp BaF2 = 1.0 x 10^-6.
Ksp BaF2 = [Ba2+(aq)]×[F-(aq)]^2 at equilibrium
When Qsp >Ksp, BaF2 will precipitate
Qsp = [Ba2+(aq)]×[F-(aq)]^2
[Ba2+(aq)]×[F-(aq)]^2 > 1.0 x 10^-6.
0.0122 moldm-3 × [F-(aq)]^2 > 1.0 x 10^-6
[F-(aq)]^2 > 1×10^-6 / 0.0122 mol2dm-6
[F-(aq)]^2 > 81.96 × 10^-6 mol2dm-6
[F- (aq)] > 9.05 × 10^-3 moldm-3
So F- concentration should be more than 9.05 × 10^-3 moldm-3
