Use your experimentally determined value of ksp and show, by calculations, that ag2cro4 should precipitate when 5ml of 0.004m agno3 are added to 5ml of 0.0024 m k2cro4.

When the value of Ksp = 3.83 x 10^-11 (should be given - missing in your Q)

So, according to the balanced equation of the reaction:

and by using ICE table:

              Ag2CrO4(s)  → 2Ag+ (Aq) + CrO4^2-(aq)

initial                                     0                   0

change                              +2X                 +X

Equ                                       2X                   X

∴ Ksp = [Ag+]^2[CrO42-]

so by substitution:

∴ 3.83 x 10^-11 = (2X)^2* X

3.83 x 10^-11 = 4 X^3

∴X = 2.1 x 10^-4 

∴[CrO42-] = X = 2.1 x 10^-4 M

[Ag+] = 2X = 2 * (2.1 x 10^-4) 

                  = 4.2 x 10^-4 M

when we comparing with the actual concentration of [Ag+] and [CrO42-]

when moles Ag+ = molarity * volume

                               = 0.004 m * 0.005L

                               = 2 x 10^-5 moles
[Ag+] = moles / total volume
     
          = 2 x 10^-5 / 0.01L

          = 0.002 M

moles CrO42- = molarity * volume

                         = 0.0024 m * 0.005 L

                         = 1.2 x 10^-5 mol

∴[CrO42-] = moles / total volume

                 = (1.2 x 10^-5)mol / 0.01 L 

                 = 0.0012 M

by comparing this values with the max concentration that is saturation in the solution 

and when the 2 values of ions concentration are >>> than the max values o the concentrations that are will be saturated.

∴ the excess will precipitate out       

Because you ate playdough