balanced net ionic equation how to do this

Question

Show the balanced net ionic equation and list spectator ions when a solution of potassium iodide and lead (II) nitrate react to produce a precipitate of lead (II) iodide suspended in a solution of potassium nitrate.

Answer 1

Lead (II) means the lead has an oxidation state of two, therefore the charge on the Lead cation is 2+
Pb ²⁺
nitrate always has a charge of -1 (you need to memorize this one)
NO₃ ^1-

So the formula for Lead (II) nitrate is
Pb(NO₃)₂ 

potassium iodide:
potassium is in group 1, all group 1 elements lose 1 electron to form cations with a charge of 1+
K+
iodine is in group 7, all group 7 elements gain 1 electron to form anions with a charge of 1-
I-

So formula for potassium iodide
KI

Balanced molecular equation

Pb(NO₃)₂(aq) + 2KI(aq) ----> PbI₂(s) + 2KNO₃(aq)

In aqueous solutions (aq) the compounds are all dissociated into their constituent ions. The Complete ionic equation shows all the ions present in solution separately.

Pb²⁺(aq) + 2NO₃^-(aq) + 2K⁺(aq) + 2I^-(aq) --> PbI₂(s) + 2K⁺(aq) + 2NO₃^- (aq)

When you look at the above equation you can see that the 2NO₃^-(aq) and the 2K⁺(aq) are the same on both sides of the equation. This means they are not participating in the reaction, they are spectator ions. The NET ionic equation shows only those species in a reaction that are actually reacting. The spectator ions are left out.

NET ionic equation:

Pb²⁺(aq) + 2I^-(aq) = PbI₂(s)