Silver iodide AgI has a Ksp value of 8.3 xx 10^-17. What is the solubility of AgI in mol/L?

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    Since you were given a ##K_sp## value which is the solubility product constant for the equilibrium of a solid with its dissociated ions we are evidently working with an equilibrium.
    Silver iodide equilibrates upon being placed into water:
    ##color(white)([( color(black)(AgI(s)) color(black)(stackrel(H_2O(l))(rightleftharpoons)) color(black)(Ag^(+)(aq)) color(black)(+) color(black)(I^(-)(aq))) (color(black)(I) color(black)(-) color(black)(0 M) color(black)(0 M)) (color(black)(C) color(black)(-) color(black)(+x) color(black)(+x)) (color(black)(E) color(black)(-) color(black)(x) color(black)(x))])##
    where ##x## is the equilibrium concentration of either ##Ag^(+)## or ##I^(-)## in solution.
    This means our equilibrium expression looks like this:
    ##K_sp = [Ag^(+)][I^(-)]##
    ##stackrel(K_sp)overbrace(8.3xx10^(-17)) = x^2##
    ##=> color(green)(x = 9.1xx10^(-9))## ##color(green)(M)##
    Since ##x## is the concentration of ##Ag^(+)## or ##I^(-)## in solution and there is a ##1:1## molar ratio of ##AgI## to either of these species…
    The molar solubility of silver iodide is ##color(blue)(9.1xx10^(-9))## ##color(blue)(M)## or ##color(blue)(mol/L)##.
    The physical interpretation of this is that silver iodide doesn’t dissociate very much. It also happens to form a yellow precipitate in solution if there isn’t enough water demonstrating its poor solubility.

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