Here’s how you can do that.
You’re dealing with a in which two soluble in aqueous solution react to form an insoluble solid that precipitates out of solution.
In this case, lead(II) nitrate, ##”Pb”(“NO”_3)_2##, and sodium iodide, ##”NaI”##, both soluble in water, will exist as ions in aqueous solution
##”Pb”(“NO”_3)_text(2(aq]) -> “Pb”_text((aq])^(2+) + 2″NO”_text(3(aq])^(-)##
##”NaI”_text((aq]) -> “Na”_text((aq])^(+) + “I”_text((aq])^(-)##
When these two are mixed, the lead(II) cations, ##”Pb”^(2+)##, and the iodide anions, ##”I”^(-)##, will bind to each other and form lead(II) iodide, an insoluble ionic compound.
The other product of the reaction is aqueous sodium nitrate, ##”NaNO”_3##, which will exist as ions in solution.
You can thus say that
##”Pb”(“NO”_3)_text(2(aq]) + color(red)(2)”NaI”_text((aq]) -> “PbI”_text(2(s]) darr + color(red)(2)”NaNO”_text(3(aq])##
The complete ionic equation, which includes all the ions present in solution, will look like this
##”Pb”_text((aq])^(2+) + 2″NO”_text(3(aq])^(-) + color(red)(2)”Na”_text((aq])^(+) + color(red)(2)”I”_text((aq])^(-) -> “PbI”_text(2(s]) darr + color(red)(2)”Na”_text((aq])^(+) + 2″NO”_text(3(aq])^(-)##
The net ionic equation, which eliminates spectator ions, i.e. the ions that are present on both sides of the equation, will look like this
##”Pb”_text((aq])^(2+) + color(red)(cancel(color(black)(2″NO”_text(3(aq])^(-)))) + color(red)(cancel(color(black)(color(red)(2)”Na”_text((aq])^(+)))) + color(red)(2)”I”_text((aq])^(-) -> “PbI”_text(2(s]) darr + color(red)(cancel(color(black)(color(red)(2)”Na”_text((aq])^(+)))) + color(red)(cancel(color(black)(2″NO”_text(3(aq])^(-))))##
This will be equivalent to
##”Pb”_text((aq])^(2+) + color(red)(2)”I”_text((aq])^(-) -> “PbI”_text(2(s]) darr##
Lead(II) iodide is a yellow solid that precipitates out of solution.