For CaCrO4, Ksp= 7.1x 10^-4. How many grams of Na2CrO4 must be added to 200 mL of .250 M Ca(NO3)2 to just initiate precipitation?

Answer 1

.......CaCrO4 ==> Ca^2+ + CrO4^2-
I......solid.......0.......0
C......solid.......x.......x
E......solid.......x.......x

........Ca(NO3)2 ==> Ca^2+ + NO3^2-
I........0.25M........0.......0
C.......-0.25.......0.25......0.25
E..........0........0.25......0.25

Note that CaCrO4 is not soluble and uses Ksp. Ca(NO3)2 is soluble so all of it has dissolved and there is no Ca(NO3)2 remaining.
Total (Ca^2+) = x from CaCrO4 and 0.25 from Ca(NO3)2

Ksp = (Ca^2+)(CrO4^2-)
You know Ksp, you know (Ca^2+) [from above it is x + 0.25), calculate (CrO4^2-) which will have units of mols/L. Also that will be (Na2CrO4) in mols/L. Convert to grams/L [g = mols x molar mass) and convert that to grams in 200 mL.
Post your work if you get stuck.