Answer:
Precipitation will occur after the concentration of fluoride ions exceeds 0.01 M.
4 mL volume of NaF must be added to cause
to precipitate.
Explanation:
Concentration of barium ions = ![[Ba^{2+}]=0.016 M](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D0.016%20M)
Volume of barium ion solution = 2.0 L

The solubility product of the barium fluoride = 
![K_{sp}=[Ba^{2+}][F^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BBa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2)
![1.6\times 10^{-6}=[0.016 M]\times [F^-]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-6%7D%3D%5B0.016%20M%5D%5Ctimes%20%5BF%5E-%5D%5E2)
![[F^-]=0.01 M](https://tex.z-dn.net/?f=%5BF%5E-%5D%3D0.01%20M)
Precipitation will occur after the concentration of fluoride ions exceeds 0.01 M.
Concentration of fluoride ion = 
Volume of solution = 
Concentration of NaF solution added = 
Volume of NaF solution added = 


1 L = 1000 mL
0.004 L = 0.004 × 1000 mL = 4 mL
4 mL volume of NaF must be added to cause
to precipitate.