[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.