Chem 2

Calculate the solubility of silver chromate, Ag2CrO4- in 0.005M Na22Cr)4- Ksp=2.6X10-12

HELP PLEASE......

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  1. Let x = solubility Ag2CrO4 at equilibrium.
    .........Ag2CrO4(s) --> 2Ag^+ + CrO4^2-
    equil.......x............2x.......x

    ..........Na2CrO4==> 2Na^+ + CrO4^2-
    initial.....0.005M.....0.......0
    change.....-0.005...2*0.005...0.005
    equil.......0........2*0.005..0.005

    Ksp = (Ag^+)^2(CrO4^2-)
    Substitute from the ICE charts above.
    (Ag^+) = 2x from Ag2CrO4.
    (CrO4^2-) = x from Ag2CrO4 + 0.005 from Na2CrO4.
    Solve for x = solubility Ag2CrO4.

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