# What is the solubility of PbI_2?

Oct 10, 2016

You need a ${K}_{\text{sp}}$ value for $P b {I}_{2}$.

#### Explanation:

Plumbous iodide is insoluble in aqueous solution, and thus the reaction would lead to a yellow precipitate of $P b {I}_{2}$ as shown.

$P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 K I \left(a q\right) \rightarrow P b {I}_{2} \left(s\right) \downarrow + 2 K N {O}_{3} \left(a q\right)$.

Now of course we could calculate the concentration of $P b {I}_{2}$ that does go up into solution, but for this need a ${K}_{\text{sp}}$ value for $P b {I}_{2}$.

$P b {I}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 {I}^{-}$

${K}_{\text{sp}} = \left[P {b}^{2 +}\right] {\left[{I}^{-}\right]}^{2}$ $=$ ??