What mass of barium fluoride, baf2, will dissolve in 500.0 ml of 0.100 m naf solution?

1) You neeed to know and use the Ksp for BaF2.

At 25°C this Ksp is 1.0 * 10 ^ - 6

2) The solutibility of BaF2 is given by:

BaF2 ⇄ Ba(2+) + 2F(-)

                x            2x

=> Ksp = x * (2x)^2 = 4x^3

3) When you have a NaF solution, you have to take into accout the concentration of the NaF solution

M = 0.1

Now the equilibrium species are:

                     BaF2 ⇄ Ba(2+)    +     2F(-)

                                       x              2x + 0.10

And Ksp = x* [2x + 0.10]^2 = 1.0 * 10 ^ -6

Given that the Ksp << 1 you may assume that 2x << 0.1 => 2x + 0.1 ≈ 0.1

=> 1.0 * 10 ^ - 6 ≈ x(0.1)^2 = 0.01x

=> x = 1.0 * 10^ -6 / 0.01 = 1.0 * 10^ - 4 M = 0.0001 M

That is the molar solubility.

4) Now, you calculate the number of moles from the molarity's formula:

M = n / v => n = M * v = 0.0001 M * 0.500 l = 0.00005 mol

And now convert to grams,

mass in grams = number of moles * molar mass

molar mass of BaF2 = 175.34 g/mol

mass in grams = 0.00005 moles * 175.34 g / mol = 0.0088 g

0.0088 g

I think it might be 159g