Question

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?

Answer 1

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.