[angel93]

so the problem is:

A solution contains 0.050 M Ba2+ and 0.040 M  Ag+. Can 99.90% of Ba2+ ion be precipitated by chromate (CrO42-) without precipitating the other metal ion?

so the Ksp(BaCrO4)= 2.1*10^-10
and Ksp(Ag2CrO4)= 1.2*10^-12

From this information I know the Ag2CrO4 will precipitate first and then I figured I would multiply 99.90% with 0.050 M to give the the concentration of of chromate precipitated. From there I am not sure what else to do.

[Borek]

If the silver precipitates first, answer to the question

Can 99.90% of Ba2+ ion be precipitated by chromate (CrO42-) without precipitating the other metal ion?

is simply "no".