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Which is the correct net ionic equation for the reaction of ammonium iodide with lead (II) nitrate to form lead (II) iodide and ammonium nitrate. You will need to first write your formula unit and total ionic equations. Recall from your first slide: Did lead (II) iodide have a high solubility or a low solubility

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Answer:

2 I⁻(aq) + Pb²⁺(aq) ⇒ PbI₂(s)

Explanation:

Let's consider the molecular equation of ammonium iodide with lead (II) nitrate to form lead (II) iodide and ammonium nitrate. Lead (II) iodide has a low solubility.

2 NH₄I(aq) + Pb(NO₃)₂(aq) ⇒ PbI₂(s) + 2 NH₄NO₃(aq)

The complete ionic equation includes all the ions and the species that do not dissociate in water.

2 NH₄⁺(aq) + 2 I⁻(aq) + Pb²⁺(aq) + 2 NO₃⁻(aq) ⇒ PbI₂(s) + 2 NH₄⁺(aq) + 2 NO₃⁻(aq)

The net ionic equation includes only the ions that participate in the reaction and the species that do not dissociate in water.

2 I⁻(aq) + Pb²⁺(aq) ⇒ PbI₂(s)

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