Answer : The full balanced ionic equation will be,
![2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)](https://tex.z-dn.net/?f=2KI%28aq%29%2BPb%28NO_3%29_2%28aq%29%5Crightarrow%202KNO_3%28aq%29%2BPbI_2%28s%29)
Reactants are lead nitrate and potassium iodide.
Products are lead iodide and potassium nitrate.
The spectator ions are, ![K^+,NO_3^-](https://tex.z-dn.net/?f=K%5E%2B%2CNO_3%5E-)
Explanation :
Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.
Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.
The full balanced ionic equation will be,
![2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)](https://tex.z-dn.net/?f=2KI%28aq%29%2BPb%28NO_3%29_2%28aq%29%5Crightarrow%202KNO_3%28aq%29%2BPbI_2%28s%29)
The ionic equation in separated aqueous solution will be,
![2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)](https://tex.z-dn.net/?f=2K%5E%2B%28aq%29%2B2I%5E%7B-%7D%28aq%29%2BPb%5E%7B2%2B%7D%28aq%29%2B2NO_3%5E%7B-%7D%28aq%29%5Crightarrow%20PbI_2%28s%29%2B2K%5E%2B%28aq%29%2B2NO_3%5E%7B-%7D%28aq%29)
In this equation,
are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
![Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)](https://tex.z-dn.net/?f=Pb%5E%7B2%2B%7D%28aq%29%2B2I%5E%7B-%7D%28aq%29%5Crightarrow%20PbI_2%28s%29)