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The reaction of aqueous cobalt(II) iodide and aqueous lead(II) nitrate is represented by the balanced formula equation.

CoI2(aq) + Pb(NO3)2(aq) → PbI2(s) + Co(NO3)2(aq)

Give the balanced ionic equation for the reaction. Include the states.

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Convert the molecular equation into the ionic equation. You do this by changing aq solution materials into ions (on both sides), then canceling any ions common to both sides.
For example, Co^+2 + 2I^- + Pb^+2 + 2NO3^- ==> PbI2(s) + Co^+2 + 2NO3^-
NOTE: Each of the ions should have (aq) after each.PbI2(s) is not an ion. It is a solid; a precipitate. Now cancel those common to both sides.
After canceling, this is what is left.
Pb^+2(aq) + 2I^-(aq) ==> PbI2(s) and that is the NET ionic equation.

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