What is the molar solubility of BaF2 in pure water?

Question

a) 3.6 x 10^-3 M

b) 3.6 x 10^-3 M

c) 2.1 x 10^-4 M

d) 4.2 x 10^-4 M

Answer 1

I have found that the product of solubility of BaF2 is 10^-6 (link)

you can write BaF2<---> Ba+2F-
if c is concentration of Ba F2

you have Ps = c * (2c)^2 = 4c^3

so c^3 = 10^-6/4= 2.510^-7

and c= 6.3 10^-3 M

Answer 2

Assume the molar solubility be ‘s’. by ionization at equilibrium( it is a sparingly soluble salt),
BaF2  Ba2+ + 2F-
At start 1 0 0 (assume)
At eq. 1-s s 2s (concentrations)

Keq = [Ba2+][F-]2
--------------
[BaF2]
Ksp=Keq[BaF2]=[Ba2+][F-]2= (s)(2s)2 = 4s3

You will find Ksp for BaF2 in any science data book( BaF2 is a sparingly soluble salt. Hence [BaF2] remains almost constant) substitute its value in the above eq and solve for ‘s’. that is the molar solubility of BaF2 in pure water.