In a solution of .200M NaI, solid AgI was placed in. Calculate the solubility if AgI (Ksp=1.5*10^-16)

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  1. Assuming complete dissociation of NaI as it is an ionic compound,

    NaI --> Na^+ + I^-
    x x (moldm^-3)
    AgI <==> Ag^+ + I^-
    y y (moldm^-3)

    y=solubility of AgI

    So the total [I^-]=(x+y) moldm^-3

    But, Ksp AgI = [Ag^+][I^-]=y(x+y) mol^2dm^-6 = 1.5*10^-16 mol^2dm^-6

    Hope this helps..

    (Please note that I haven't mentioned the state of the above compounds)

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  2. Thank you so much!

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