# Question #307a5

Jun 8, 2014

${\text{PbI}}_{2}$ would not form ions in the ionic equation.

#### Explanation:

$P b {I}_{2}$ would not form ions in the ionic equation.

The ionic equation for the example above is $P {b}^{2 +} \left(a q\right) + 2 {I}^{-} \left(a q\right) \to P b {I}_{2} \left(s\right)$

The way to reduce a full balanced symbol equation to an ionic equation is to remember that when in solution, the ions present are free to move around independently therefore anything with an (aq) state symbol can be written as independent ions rather than as a compound:

Like so: $P {b}^{2 +} \left(a q\right) + 2 N {O}_{3}^{-} \left(a q\right) + 2 {K}^{+} \left(a q\right) + 2 {I}^{-} \left(a q\right)$
$\to P b {I}_{2} \left(s\right) + 2 {K}^{+} \left(a q\right) + 2 N {O}_{3}^{-} \left(a q\right)$

Now cancel any spectator ions - those which are the same, with the same state symbol, on both sides of the equation, and you have the ionic equation.