May 22, 2015

You can add a solution of potassium iodide, $K I$.

The balanced chemical equation for this double replacement reaction would look like this

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

Lead (II) iodide is a yellow precipitate that will be formed in the reaction.

The ney ionic equation, which you get if the spectator ions are removed, looks like this

$P {b}_{\left(a q\right)}^{2 +} + 2 {I}_{\left(a q\right)}^{-} \to P b {I}_{2 \left(s\right)} \downarrow$

Here's a video of the reaction: