Calculate the molar solubility of barium fluoride in each of

Calculate the molar solubility of barium fluoride in each of the following.

Solution

)in 0.12 M Ba(NO3)2...
Ba(NO3)2 dissociates completely ,therefore in 0.12 M Ba(NO3)2 solution ...[Ba2+] = 0.12 M
so now concentration of Ba(2+) will be (s+ 0.12) moles/litre or 0.12 moles/litre approximately ..
so again using the same expression...
Ksp = [Ba2+] [F-]^2
putting the values..
2.45 X 10^-5 = 0.12 X (2s)^2
2.45 X 10^-5 = 0.12 X 4s^2
2.45 X 10^-5 = 0.48 X s^2
5.104 X 10^-5 = s^2
s = 7.144 X 10^-3 mole/L

C) similarly in 0.15 M NaF ...concetration of F(-) will be (0.15 + 2s) moles/L or 0.15 moles/L approximately..
so as Ksp = [Ba2+] [F-]^2
2.45 X 10^-5 = s X (0.15)^2
2.45 X 10^-5 = s X 0.0225
s = 2.45 X 10^-5/0.0225 = 1.09 X 10^-3 moles/L