The reaction between sodium iodide (NaI) and lead(II) acetate (Pb(C2H3O2)2) is a double displacement reaction that results in the formation of a precipitate.

Here's the balanced chemical equation:

2 NaI (aq) + Pb(C2H3O2)2 (aq) → PbI2 (s) + 2 NaC2H3O2 (aq)

Here's what happens:

* NaI and Pb(C2H3O2)2 are both soluble ionic compounds, meaning they dissociate into ions when dissolved in water.

* The lead(II) ions (Pb²⁺) from lead(II) acetate react with the iodide ions (I⁻) from sodium iodide to form lead(II) iodide (PbI2), which is a bright yellow insoluble solid (precipitate).

* The remaining ions, sodium ions (Na⁺) and acetate ions (C2H3O2⁻), stay in solution as sodium acetate (NaC2H3O2), a soluble ionic compound.

Therefore, the solution for NaI plus PbC2H3O2 is the formation of a yellow precipitate of lead(II) iodide (PbI2) and a solution of sodium acetate (NaC2H3O2).