Calculate the molar solubility of SrF2 in 0.015 M of NaF.

Question

Ksp for SrF2 = 4.33 x 10^ (-9)

I think I am solving the problem correctly but I can't figure out the math at the end to solve for x. Here's what I've done so far.

from SrF2 <---> Sr^2 2F

Ksp = (x) (2x 0.015)^2

4.33 x 10^ (-9) = (x) (2x 0.015)^2

4.33 x 10^ (-9) = x(4x^2 0.06x 0.000 225)

4.33 x 10^ (-9) = 4x^3 0.06x^2 0.000 225x

Answer 1

SrF2 &lt;---&gt; Sr+2    &amp; 2F-

K = [Sr] [F]^2

4.33 e-9  = [Sr] [0.15]^2

4.33 e-9  = [Sr] (0.0225)

[Sr] = 4.33e-9 / 0.0225

Sr = 1.92e-7 molar

that amount of Sr+2 shows up in the solution because that amount of SrF2 &quot;dissolved&quot; into the solution

your answer: the molar solubility of SrF2 is 1.9e-7