0.01 mole of AgNO3 is added to 1 litre of a solution which is 0.1 M in Na2CrO4 and 0.005 M in NaIO3. Calculate the millimol of precipitate formed at equilibrium.Ksp values of Ag2CrO4 and AgIO3 are 10-8 and 10-13 respectively).

The Ksp values of Ag2CrO4 and AgIO3 reveals that CrO42- and IO3-will be  precipitated on addition of AgNO3 as: [Ag+][IO3−]=  10−13[Ag+]needed=10−13[0.005]=2×10−11[Ag+]2[CrO42−]= 10−8[Ag+]needed=10−80.1=3.16×10−4Thus, AgIO3 will be precipitate first.Now, in order to precipitate AgNO3 one can show:AgNO30.01 0.005+NaIO30.005     0→ AgIO3   0 0.005+NaNO3  0 0.005The left mole of AgNO3 are now used to precipitate Ag2CrO42AgNO30.005     0+Na2CrO4   0.1 0.0975→ Ag2CrO4   0 0.0025+2NaNO3  0 0.005