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”_2″O”(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.
