How do you write the the reaction of lead(II) nitrate (aq) with sodium iodide (aq) to form lead (II) iodide precipitate and sodium nitrate solution?

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.

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