Question

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

Answer 1

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.