Ba2 2F BaF2 Kc 188 x 10 In the solution at equilibrium th

Ba2+ + 2F- --> BaF2 Kc = 1.88 x 10 In the solution at equilibrium, the F- concentration was 0.250M. Determine the equilibrium concentration of the Ba2+ ion.

Kc = [BaF2-]/[Ba2+][F-]^2 , Kc for reverse rxn = [F-]^2[Ba2+]/[BaF2] = (1/1.88x10) = 0.0532,

if initially only BaF2 is present then , at equi we have [F-] = 0.25, [Ba2+] = 2x 0.25 = 0.5M