Question

What is the net ionic equation for the reaction of aqueous lead(II) nitrate with aqueous sodium bromide?

Answer 1

`"Pb"_ ((aq))^(2+) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr`

Explanation:

Lead(II) nitrate, `"Pb"("NO"_3)_2`, and sodium bromide, `"NaBr"`, are soluble in aqueous solution, which means that they dissociate completely to form cations and anions when dissolved in water.

`"Pb"("NO" _ 3)_ (2(aq)) -> "Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)`

`"NaBr"_ ((aq)) -> "Na"_ ((aq))^(+) +"Br"_ ((aq))^(-)`

When these two solutions are mixed, lead(II) bromide, `"PbBr"_2`, an insoluble ionic compound, and sodium nitrate, `"NaNO"_3`, another soluble ionic compound, will be formed.

The complete ionic equation will look like this

`"Pb"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2 xx ["Na"_ ((aq))^(+) +"Br"_ ((aq))^(-)] -> "PbBr"_ (2(s)) darr + 2xx ["Na" _ ((aq))^(+) + "NO"_ (3(aq))^(-)]`

Notice that the nitrate anions, `"NO"_3^(-)`, and the sodium cations, `"Na"^(+)`, exist as ions on both sides of the equation.

This tells you that these two chemical species act as spectator ions. In order to write the net ionic equation, you remove the spectator ions and focus on the insoluble solid and the ions that form it

`"Pb"_ ((aq))^(2+) + color(blue)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-)))) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr + color(blue)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(Black)(2"NO"_ (3(aq))^(-))))`

The net ionic equation fo this reaction will thus be

`"Pb"_ ((aq))^(2+) + 2"Br"_ ((aq))^(-) -> "PbBr"_ (2(s)) darr`

Lead(II) bromide is a white insoluble ionic compound that precipitates out of solution.