Question #60338

1 Answer
Jul 1, 2015

The answer is E) All the above ions are present in the complete ionic equation.


The complete ionic equation shows all the ions that exist when two solutions that contain ionic compounds are mixed together, regardless if they actually take part in the reaction or not.

In your case, lead (II) nitrate and potassium iodide are both soluble compounds that exist as ions in solution

#Pb(NO_3)_(2(aq)) -> Pb_text((aq])^(2+) + 2NO_(3(aq))^(-)#


#KI_((aq)) -> K_((aq))^(+) + I_((aq))^(-)#

When you mix two solutions that contain these ions together, lead iodide, a yellow precipitate, is formed.

The ions that don't become a part of the insoluble solid will remain in solution and act as spectator ions.

So, if you put all this together, you'll get

#Pb_text((aq])^(2+) + 2NO_(3(aq))^(-) + 2K_((aq))^(+) + 2I_((aq))^(-) -> PbI_(2(s)) darr + 2K_((aq))^(+) + 2NO_(3(aq))^(-)#

As you can see, all the ions listed in your question are a part of the complete ionic equation.

The net ionic equation will be

#Pb_text((aq])^(2+) + 2I_((aq))^(-) -> PbI_(2(s)) darr#