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Chemistry

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?

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Cory

  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.

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    DrBob222

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