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QUESTION

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

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.