The reaction of aqueous sodium bromide and aqueous lead(II) nitrate is represented by the balanced net ionic equation.

2Br-(aq) + Pb2+(aq) → PbBr2(s)

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

This question is asking for the reverse of your previous posts; that is, the question wants you to take the net ionic equation given and make the molecular equation, then the ionic equation from that.

2NaBr(aq) + Pb(NO3)2(aq) ==> PbBr2(s) + 2NaNO3(aq), then
2Na^+(aq) + 2Br^-(aq) + Pb^+2(aq) + 2NO3^-(aq) ==> PbBr2(s) + 2Na^+2(aq) + 2NO3^-(aq)
Check my work to make sure all of the charges and (aq) marks are showing correctly.

To write the balanced ionic equation for the reaction between aqueous sodium bromide and aqueous lead(II) nitrate, let's start by writing the complete molecular equation:

2NaBr(aq) + Pb(NO3)2(aq) → PbBr2(s) + 2NaNO3(aq)

Next, let's separate the soluble compounds into their ionic forms, while indicating their states:

2Na+(aq) + 2Br-(aq) + Pb2+(aq) + 2NO3-(aq) → PbBr2(s) + 2Na+(aq) + 2NO3-(aq)

Since the Na+(aq) and NO3-(aq) ions appear on both sides of the equation, they can be canceled out. Therefore, the balanced ionic equation for the reaction is:

2Br-(aq) + Pb2+(aq) → PbBr2(s)

To write the balanced ionic equation for the reaction between aqueous sodium bromide (NaBr) and aqueous lead(II) nitrate (Pb(NO3)2), you need to first write the balanced molecular equation and then convert it into the ionic equation.

1. Write the balanced molecular equation:
NaBr(aq) + Pb(NO3)2(aq) → PbBr2(s) + 2NaNO3(aq)

2. Split the reactants and products into their respective ions:
Na+(aq) + Br-(aq) + Pb2+(aq) + 2NO3-(aq) → PbBr2(s) + 2Na+(aq) + 2NO3-(aq)

3. Cancel out the spectator ions (ions that appear on both sides of the equation without undergoing any change):
In this case, Na+ and NO3- are spectator ions because they appear on both sides of the equation. Canceling them out leaves us with:
Br-(aq) + Pb2+(aq) → PbBr2(s)

Therefore, the balanced ionic equation for the reaction is:
2Br-(aq) + Pb2+(aq) → PbBr2(s)

Make sure to include the states of each species to indicate whether they are in a solid (s), liquid (l), gas (g), or aqueous (aq) state. In this case, all the reactants and the product are in their aqueous form (aq), except for PbBr2, which is a solid precipitate (s).