Question

1. build the net ionic equation for the reaction that produces the lead(ll) iodide precipitate.

a net ionic equation shows only the ions that participate in the reaction. these are the ions that react to form the precipitate.
ions that remain aqueous in a reaction are called spectators. the spectator ions are not included in the net ionic equation.
what is a net ionic equation?
cation + anion → precipitate
choices:
i₂ pbi₂ 2 pb 1 pb²⁺ i⁻

Explanation:

Step1: Identify the cation and anion for lead(II) iodide

The precipitate is lead(II) iodide, \( \text{PbI}_2 \). The cation in lead(II) iodide is \( \text{Pb}^{2+} \) (lead(II) ion) and the anion is \( \text{I}^- \) (iodide ion).

Step2: Determine the stoichiometry

To form \( \text{PbI}_2 \), we need 1 \( \text{Pb}^{2+} \) ion and 2 \( \text{I}^- \) ions because the formula of lead(II) iodide is \( \text{PbI}_2 \) (the charge of \( \text{Pb}^{2+} \) is +2 and each \( \text{I}^- \) is -1, so 2 \( \text{I}^- \) ions are needed to balance the charge: \( (+2) + 2\times(-1) = 0 \)).

Step3: Write the net ionic equation

The net ionic equation is \( \text{Pb}^{2+} + 2\text{I}^-
ightarrow \text{PbI}_2 \). So the coefficients and ions are: 1 (for \( \text{Pb}^{2+} \)), 2 (for \( \text{I}^- \)), and 1 (for \( \text{PbI}_2 \)).

Answer:

1 \( \text{Pb}^{2+} \) + 2 \( \text{I}^- \) \(
ightarrow \) 1 \( \text{PbI}_2 \)