Question

A saturated solution of barium fluoride, BaF2, was prepared by dissolving solid BaF2 in water. The concentration of Ba2+ ion in the solution was found to be 7.52

Answer 1

There are 2 questions in this and I have answered both.

Answer 1:
Let solubility be x
                                     BaF2 <---> Ba2+   + 2F-
at equilibrium:                             x              2x
given:[Ba2+]=7.52*10^-3 M =x
so, [F-]=2*x = 2* 7.52*10^-3 = 0.01504
Ksp = [Ba2+][F-]^2
        = (7.52*10^-3 ) (0.01504)^2
        = 1.7*10^-6
Answer 2:
Let solubility be x
                                         Ag2CO3<--->Ag+   + CO32-
at equilibrium:                                     x               x
Ksp = [Ag+] [CO3 2-]
8.1*10^-12 = x*x
x= 2.85*10^-6 mol/L
molar mass of Ag2Co3 = 276 gm
solubility in terms of moL/l = 2.85*10^-6 mol/L
but 1 mol = 276 gm
solubility in terms of gm/l = 2.85*10^-6 *276 = 7.9*10^-4 gm/L
Ans: solubility in terms of gm/L=7.9*10^-4 gm/L