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.

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.

Co2+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq) → PbI2(s) + Co2+(aq) + 2NO3-(aq)

Well, this equation seems to be having quite the party, doesn't it? We've got some cobalt(II) ions, iodide ions, lead(II) ions, and nitrate ions all mingling together. They're just having a ball, dancing around and swapping partners.

But things take a turn when lead(II) bromide, a rather shy compound, decides to join the party. Suddenly, a beautiful dance number ensues, with the lead(II) ions and iodide ions coming together to form solid lead(II) iodide. Meanwhile, the cobalt(II) ions, being the life of the party, refuse to settle down and continue to dance around with the nitrate ions.

So, here's the balanced ionic equation for this chaotic dance party:

Co2+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq) → PbI2(s) + Co2+(aq) + 2NO3-(aq)

I hope you enjoyed that wild dance-off between the ions!

To write the balanced ionic equation, we need to separate the soluble ionic compounds into their constituent ions. The insoluble compound, PbI2, will remain intact since it does not dissociate.

The given balanced formula equation is:
CoI2(aq) + Pb(NO3)2(aq) → PbI2(s) + Co(NO3)2(aq)

Now let's write the balanced ionic equation including the states:

Co2+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq) → PbI2(s) + Co2+(aq) + 2NO3-(aq)

Note that the Co2+ and NO3- ions are present on both sides of the equation, so they cancel out. The balanced ionic equation is:

2I-(aq) + Pb2+(aq) → PbI2(s)

This equation represents the net ionic equation where only the species actively participating in the reaction are shown.

To write the balanced ionic equation for a reaction, we need to break down the reactants and products into their respective ions. Here's how you can do it:

1. Start by writing the molecular equation for the reaction:
CoI2(aq) + Pb(NO3)2(aq) → PbI2(s) + Co(NO3)2(aq)

2. Write the complete and balanced ionic equation by separating the reactants and products into their respective ions:
Co2+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq) → PbI2(s) + Co2+(aq) + 2NO3-(aq)

3. Notice that there are Co2+ ions and NO3- ions on both sides of the equation. These ions are called spectator ions because they do not participate in the actual chemical reaction. Therefore, we can eliminate them from the equation to simplify it:

2I-(aq) + Pb2+(aq) → PbI2(s)

This is the balanced ionic equation for the reaction. Note that we removed the spectator ions Co2+ and NO3-. The state symbols indicating the states of matter (aq for aqueous and s for solid) are also included in the equation.

So, the balanced ionic equation is:
2I-(aq) + Pb2+(aq) → PbI2(s)