Question

Calculate the molar solubility of CaF2 in a solution containing 0.325 M of Ca(NO3)2. The Ksp value for CaF2 is 1.46×10−10. can you also explain me how to solve for X toward the end of the problem

Answer 1

CaF2(s) <-----> Ca+2(aq) + 2F-(aq)

Ca(NO3)2 -----> Ca+2(aq) + NO3-(aq)

[Ca+2] = S = 0.325 M

kSP of CaF2 = (s)(2s)^2

   3.9*10^-11 = 0.325*(2s)^2

    s = molar solubility of CaF2 = 5.48*10^-6 mol/L