reaction of lead nitrate with sodium iodide
2 NaI + Pb(NO3)2 -> PbI2 + 2 NaNO3
The PbI2 will precipitate out.
The reaction between lead nitrate (Pb(NO3)2) and sodium iodide (NaI) is a double replacement or precipitation reaction. During the reaction, the lead cations (Pb2+) from lead nitrate combine with the iodide anions (I-) from sodium iodide to form lead iodide (PbI2), while sodium cations (Na+) combine with nitrate anions (NO3-) to form sodium nitrate (NaNO3). The balanced chemical equation for the reaction is:
Pb(NO3)2 + 2NaI → PbI2 + 2NaNO3
To determine the reaction and products, you need to identify the ions involved and their charges.
In lead nitrate (Pb(NO3)2), the lead ion has a 2+ charge (Pb2+) while the nitrate ion has a 1- charge (NO3-).
In sodium iodide (NaI), the sodium ion has a 1+ charge (Na+) and the iodide ion has a 1- charge (I-).
To find out the products, you swap the anions between the reactants while keeping the cations the same. In this case, the lead cation combines with the iodide anion to form lead iodide (PbI2). The sodium cation combines with the nitrate anion to form sodium nitrate (NaNO3).
Therefore, when lead nitrate reacts with sodium iodide, the products formed are lead iodide (PbI2) and sodium nitrate (NaNO3).