Chemistry solubility and precipitate

[temari]

Homework Statement

A solution contains 1.73 x 10^-3 mol/L of Ba(NO3)2(aq) and 1.2 x 10^-3 moles of NaF. What mass of KF added to this solution would produce 2.7 x 10^-3g of precipitate from this solution if the Ksp of BaF2 is
1.84 x 10-7 and you have 1.0L of solution?

Homework Equations

The Attempt at a Solution

Mass of BaF2 = 2.7*10^-3g/ 175.3 (molar mass)
= 1.5*10^-5 moles/L

Ksp = (ba) (F)^2
1.84*10^-7 = (1.73 x 10^-3 - 1.5*10^-5) (Concentration of F)^2
(1.84*10^-7 /.001715)^.5 = (Concentration of F)
.0103 = (Concentration of F at Saturation Point)

2 x 10^-3 moles of NaF + (Added Concentration of KF) - 2(1.5*10^-5 moles/L) = .0103

(Added Concentration of KF) = .00833 mol/L
.00833 mol/L * (Molar Mass of KF 58.1) = .484 g of KF

Did I answer this Question Correctly

 

[Borek]

I have not checked the answer, but the approach looks OK.