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# 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?

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Mar 20, 2016

Here's how you can do that.

#### Explanation:

You're dealing with a double replacement reaction in which two soluble ionic compounds 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, $\text{NaI}$, both soluble in water, will exist as ions in aqueous solution

${\text{Pb"("NO"_3)_text(2(aq]) -> "Pb"_text((aq])^(2+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

${\text{NaI"_text((aq]) -> "Na"_text((aq])^(+) + "I}}_{\textrm{\left(a q\right]}}^{-}$

When these two solutions are mixed, the lead(II) cations, ${\text{Pb}}^{2 +}$, and the iodide anions, ${\text{I}}^{-}$, will bind to each other and form lead(II) iodide, an insoluble ionic compound.

The other product of the reaction is aqueous sodium nitrate, ${\text{NaNO}}_{3}$, which will exist as ions in solution.

You can thus say that

${\text{Pb"("NO"_3)_text(2(aq]) + color(red)(2)"NaI"_text((aq]) -> "PbI"_text(2(s]) darr + color(red)(2)"NaNO}}_{\textrm{3 \left(a q\right]}}$

The complete ionic equation, which includes all the ions present in solution, will look like this

${\text{Pb"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) + color(red)(2)"Na"_text((aq])^(+) + color(red)(2)"I"_text((aq])^(-) -> "PbI"_text(2(s]) darr + color(red)(2)"Na"_text((aq])^(+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

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

${\text{Pb"_text((aq])^(2+) + color(red)(2)"I"_text((aq])^(-) -> "PbI}}_{\textrm{2 \left(s\right]}} \downarrow$

Lead(II) iodide is a yellow solid that precipitates out of solution.

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