Answer to Question #157642 in Inorganic Chemistry for Siluni Chandrasiri

Question #157642

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


1
Expert's answer
2021-01-25T04:23:34-0500

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.



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