When 20.0 mL of 0.100 M AgNO3 are mixed with 80.0 mL of 0.0100 M Na2CrO4, what is the chromate ion concentration (CrO4^2-)?

Ksp for Ag2CrO4 = 9.0 x 10^-12.

I think it's 1.0 x 10^-9. Am I right?

2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NaNO3

I see it differently.
Initial:
20 mL x 0.1 M = 2 millimoles AgNO3
80 mL x 0.01 M = 0.8 mmoles Na2CrO4.
Ag2CrO4 = O
NaNO3 = 0

final:
Ag2CrO4(s) = ppt = 0.8 mmoles.
Na2CrO4 = 0 (all of it used--the only chromate from the solution will come from the solubility of the silver chromate).
AgNO3 = 2-(0.8x2) = 0.4 mmoles
NaNO3 = 2 x 0.8 = 1.6 mmoles.

final concns:
Ag2CrO4 (not applicable since it is a ppt).
Na2CrO4 = 0
AgNO3 = 0.4 mmoles/100 mL = 0.004 M
NaNO3--not needed.

Ag2CrO4 ==> 2Ag^+ + CrO4^-2
Ksp + (Ag^+)^2(CrO4^-2) = 9.0 x 10^-12
The AgNO3 left unreacted is a common ion to the ppt of Ag2CrO4. By Le Chatelier's principle the Ag2CrO4 is made less soluble by the presence of excess Ag^+. Solve Ksp for (CrO4^-2) = Ksp/(Ag^+)^2 = 9.0 x 10^-12/(0.004)^2 = not 1 x 10^-9.

To find the chromate ion concentration (CrO4^2-) in the solution after mixing AgNO3 and Na2CrO4, we can use the concept of stoichiometry and the known equilibrium constant (Ksp) for Ag2CrO4.

First, let's calculate the number of moles of AgNO3 and Na2CrO4 used:

Number of moles of AgNO3 = volume (in L) x molarity
= 20.0 mL * (1 L / 1000 mL) * 0.100 M
= 0.002 mol

Number of moles of Na2CrO4 = volume (in L) x molarity
= 80.0 mL * (1 L / 1000 mL) * 0.0100 M
= 0.008 mol

Since the ratio between AgNO3 and Ag2CrO4 is 1:1, the number of moles of Ag2CrO4 formed will also be 0.002 mol.

Now, we need to determine the concentration of CrO4^2-. Since one mole of Ag2CrO4 forms two moles of CrO4^2- ions, the concentration of CrO4^2- will be half the concentration of Ag2CrO4.

Concentration of CrO4^2- = (0.002 mol / (20 mL + 80 mL)) / 2
= (0.002 mol / 0.100 L) / 2
= 0.01 M / 2
= 0.005 M

So, the chromate ion concentration (CrO4^2-) is 0.005 M.

Therefore, your initial intuition that the chromate ion concentration is 1.0 x 10^-9 M is incorrect. The correct answer is 0.005 M.