Precipitate question?

Question

if 5.0 * 10^-4 moles of Ba(No3)2 were added to 1.0 L of a
6.0 *10^-3 M solution of NaF, would a precipitate form? Show calculation

Answer 1

NaNO3 is very soluble
BaF2 has Ksp=1.84*10^-7

BaF2 <=> Ba(+2) + 2F(-)

so Ksp=[Ba(+2)][F(-)]^2 This is for the concentrations at equilibrium.
If the product P=[Ba(+2)][F(-)]^2 for the initial concentrations (and not the concentrations at equilibrium) is P>Ksp then you have a precipitate, if P=Ksp then you have a saturated solution and if P
[Ba+2]=mole/V= 5*10^-4/1= 5*10^-4

NaF -> Na(+) + F(-)
Therefore [F(-)]=[NaF]=6*10^-3

P=[Ba(+2)][F(-)]^2 = (5*10^-4)*(6*10^-3)^2 =1.80*10^-8

Answer 2

Okay, first, if I am correct, you do not need to know about how much there is of each compound. But there are solubility rules to tell you which will form a solid. SInce the equation changes, the compounds switch so it becomes, BaF and NNo3 (not balanced)
The rules state that nitrate is soluable and that means that it stays aqueous and does not form a solid. Fluoride, however, has an exception with Barium, when it is with Barium it forms a solid. These rules were in my Chem book- if you don't have them you should copy them because they are really useful