Question

What is the net ionic equation for the reaction between 2 equiv sodium chloride and 1 equiv lead acetate?

Answer 1

A net ionic equation includes only the ions that reacted and their product.

Explanation:

Balanced molecular equation:

`"2NaCl(aq) + Pb(C"_2"H"_3"O"_2)_2"` `rarr` `"PbCl"_2("s") + "2NaC"_2"H"_3"O"_2"`

Separate the individual ions in each aqueous ionic compound, and add the precipitate (solid) as a complete formula.

Complete Ionic equation:

`"2Na"^(+)("aq") + "2Cl"^(-)("aq") + "Pb"^(2+)("aq") + "2C"_2"H"_3"O"_2^(-)("aq")"` `rarr` `"PbCl"_2("s") + "2Na"^(+)("aq") + "2C"_2"H"_3"O"_2^(-)("aq")`

The `"Na"^(+)` and `"C"_2"H"_3"O"_2^(-)` are spectator ions because they do not participate in the reaction.

Net ionic equation:

Remove the spectator ions and Include only the ions that reacted and the precipitate.

`"2Cl"^(-)("aq") + "Pb"^(2+)("aq")` `rarr` `"PbCl"_2("s")"`

Answer 2

Well, all of the ions will appear in the complete ionic equation....

Explanation:

In the `"net ionic equation"` ONLY those ions that undergo macroscopic change will appear.....and thus.....

`Pb^(2+) +2Cl^(-) rarr PbCl_2(s)darr`

Lead chloride is a white salt, that is fairly insoluble in aqueous solution, and hence it `"precipitates....."` Acetate, and sodium ions are along for the ride.......i.e.

`Pb(OAc)_2(aq) +2NaCl(aq) rarr PbCl_2(s)darr+2NaOAc(aq)`....