Verified answer

Let 'x' represent the concentration of Ba2+.

|||||||||||||BaF2| Ba2+ | F-

Initial||||||||| - | 0 | 0.110 M

change|||| - | +x | +2x

Final|||||||||| - | x | 0.110 M + 2x

(Sorry if the columns in the above table don't line up properly.

The initial [F-] is 0.110 M because that is the [NaF])

Ksp = [Ba2+] * [F-]^2

1.80 * 10^-7 = x * (0.110 + 2x)^2

Since Ksp is small, assume x << 0.110 M (x is much smaller than 0.110 M)

Therefore (0.110 + 2x) equals approximately 0.110

So,

1.80 * 10^-7 = x * (0.110)^2

solve

x = 1.488 * 10^-5 M = [Ba2+]

To check the approximation:

1.80 * 10^-7 = x * (0.110 + 2x)^2 (from above)

Left side = 1.80 * 10^-7

Right side = x * (0.110 + 2x)^2

Right side = 1.488 * 10^-5 * (0.110 + 2*1.488 * 10^-5)^2

Right side = 1.80 * 10^ -7

So Left side = Right side

This means it is a good approximation.

Back to the problem:

moles of Ba2+ = [Ba2+] * Volume

moles of Ba2+ = 1.488 * 10^-5 mol/L * 0.488 L

moles of Ba2+ = 7.260 * 10^-6 mol

For every mol of Ba2+, there is 1 mol of BaF2, so

moles of BaF2 = moles of Ba2+

moles of BaF2 = 7.260 * 10^-6 mol

mass of BaF2 = moles of BaF2 * molar mass of BaF2

mass of BaF2 = 7.260 * 10^-6 mol * 175.4 g/mol

mass of BaF2 = 1.27 * 10^-3 g

Therefore 1.27 * 10^-3 g of BaF2 will dissolve.