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chemistry

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The solubility of the fictitious compound, administratium fluoride (AdF3) in water is 3.091×10−4 M. Calculate the value of the solubility product Ksp.

  • chemistry - ,

    ........AdF3 ==> Ad^3+ + 3F^-
    I.......solid.....0.......0
    C.....x dissolves..x......3x
    E.......solid......x......3x

    Ksp = (Ad^3+)(F^-)^3
    Ksp = (x)(3x)^3
    You know x = 3.091E-4. Substitute and solve for Ksp.

  • chemistry - ,

    2.46x10^-13

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