Question

Barium nitrate is added to a solution of .025 M sodium fluoride.

a) At what concentration of Ba+2 does a precipitate start to form?
b) Enough barium nitrate is added to make [Ba+2]= .0045 M. What percentage of the original fluoride ion has precipitated?

I'm having a lot of trouble writing an equilibrium equation for some reason.
Would it be something along the lines of:
Ba(NO3)2 + 2NaF(aq) <--> 2NaNO3 + Ba+2 + 2F-
?

BaF2 is a precipitate. So a molecular equation would be
Ba(NO3)2 + 2NaF ==>2Na^+ + 2NO3^- + BaF2

For the solubility of BaF2,
BaF2(s) ==> Ba^+2 + 2F^-
Ksp = (Ba^+2)(F^-)^2

(a) You know F ion. It is 0.025M. Plug into Ksp and solve for Ba ion.

(b) Knowing Ba ion is 0.0045M, solve for F ion. That subtracted from the inbitial amount will be the about pptd. Then take the percent of that. Post your work if you get stuck.