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

Jun 21, 2016

Since you were given a ${K}_{\text{sp}}$ 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:

$\textcolor{w h i t e}{\left[\begin{matrix}\text{ & color(black)("AgI"(s)) & color(black)(stackrel("H"_2"O"(l))(rightleftharpoons)) & color(black)("Ag"^(+)(aq)) & color(black)(+) & color(black)("I"^(-)(aq)) \\ color(black)("I") & color(black)(-) & "" & color(black)("0 M") & "" & color(black)("0 M") \\ color(black)("C") & color(black)(-) & "" & color(black)(+x) & "" & color(black)(+x) \\ color(black)("E") & color(black)(-) & "" & color(black)(x) & } & \textcolor{b l a c k}{x}\end{matrix}\right]}$

where $x$ is the equilibrium concentration of either ${\text{Ag}}^{+}$ or ${\text{I}}^{-}$ in solution.

This means our equilibrium expression looks like this:

K_"sp" = ["Ag"^(+)]["I"^(-)]

$\stackrel{{K}_{\text{sp}}}{\overbrace{8.3 \times {10}^{- 17}}} = {x}^{2}$

$\implies \textcolor{g r e e n}{x = 9.1 \times {10}^{- 9}}$ $\textcolor{g r e e n}{\text{M}}$

Since $x$ is the concentration of ${\text{Ag}}^{+}$ or ${\text{I}}^{-}$ in solution, and there is a $1 : 1$ molar ratio of $\text{AgI}$ to either of these species...

The molar solubility of silver iodide is $\textcolor{b l u e}{9.1 \times {10}^{- 9}}$ $\textcolor{b l u e}{\text{M}}$, or $\textcolor{b l u e}{\text{mol/L}}$.

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.