# Question #3c1d8

Jun 29, 2015

The answer is D) $P {b}^{2 +}$, ${I}^{-}$.

#### Explanation:

The double replacement reaction that takes place in aqueous solution between lead (II) nitrate, $P b {\left(N {O}_{3}\right)}_{2}$, and potassium iodie, $K I$, forms an Insoluble solid, lead (II) iodide, which precipitates out of the solution.

The other product of this reaction is potassium nitrate, $K N {O}_{3}$, which is a soluble salt.

This means that you can determine which ions will act as spectator ions by looking at what the reaction forms. Right from the start, knowing that lead (II) iodide is an insoluble solid, the ions that come together to form this compound cannot be spectator ions.

So the two candidates that stand out are $P {b}^{2 +}$ and ${I}^{-}$.

To double-check, write the complete ionic equation for this reaction

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

As you can see, the potassium and nitrate ions are present on both sides of the equation, which means that they are indeed spectator ions.

The $P {b}^{2 +}$ and ${I}^{-}$ ions are only present as free ions on the reactants' side, since they are stuck together in the insoluble solid on the products' side.

Therefore, you have

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

The net ionic equation is

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

Here's how the precipitate looks like