Ksp: What am I doing wrong?

Question

How many grams of BaF2 will dissolve in .480 L of a .110 M NaF solution?

BaF2 -> [Ba] [F]^2

Initial 0 .110
Change x 2x
Eq x .110 +2x

Assume x is negligible.

Ksp= 1.80e-7 = (2x)^2 (.110)

(4x)^2(.110) = 1.80e-7

(1.80e-7) / (.110) = 1.636e-6 4x^2

(1.636e-4) / (4) = 4.09e-7 x^2

x= square root of (4.09e-7)

x= 6.395e-4

MM= 175.324

(175.324) (6.395e-4) = .11211 g/L

(.11211 g/L) (.480 L) = .05381745 g ?

Answer 1

.BaF2..-->Ba2+.+...2F-
I..................0.........0.11...
C................x.........2x....
E...............x.....0.11-2x
Thus,

Ksp = x(0.11-2x)^2 = 1.8E-7, assume x is small, 0.11-2x= 0.11,

ksp = x*(0.11)^2 = 1.8E-7, solve for x, that is your equilibrium conc of Ba2+. I believe you can continue at this point.