[Ag+]= (.001L)(o.500mol)/(0.002L)=0.250M

This is the concentration of Ag

^{+} in the mixed solution

before precipitation. You need the concentration left in solution

after precipitation has occurred. You could set up a quadratic equation, but because the solubility of AgCl is very low you can simplify the problem by assuming that virtually all the Cl

^{-} is precipitated as AgCl.

How many moles of Cl

^{-} is this?

So how many moles of Ag

^{+} are precipitated, and how many moles are left in solution?

Using K

_{sp}, what is the concentration of Cl

^{-}? Verify that this is a very small fraction of the original Cl

^{-}, so that the simplification is justified.