Answer to Question #157642 in Inorganic Chemistry for Siluni Chandrasiri

Question

Reaction between lead(ii) nitrate and sodium iodide to produce lead(ii) iodide and sodium nitrate

Answer 1

Pb(NO₃)₂ (aq) + 2 NaI (aq) → PbI₂ (s) + 2 NaNO₃ (aq)

2Na(+) +2I(-) + Pb(2+) + 2NO3(-) = 2Na(+) +PbI2(s) + 2NO3(-)

2I(-) + Pb(2+) = PbI2 (s)

Lead iodide appears as a yellow crystalline solid. Insoluble in water.