How do you write a net ionic equation? Show an example please.

1 Answer
May 18, 2017

I am pretending you don't know how to write the complete ionic reaction either, so let's just start from there.

Let's say we have silver nitrate reacting with sodium phosphate. Then, our unbalanced reaction is:

#"AgNO"_3(aq) + "Na"_3"PO"_4(aq) -> "Ag"_3"PO"_4(s) + "NaNO"_3(aq)#

To balance this, one way is to balance the sodium cations, and then the nitrate polyatomic ions, which consequently also balances the silver cations.

#3"AgNO"_3(aq) + "Na"_3"PO"_4(aq) -> "Ag"_3"PO"_4(s) + 3"NaNO"_3(aq)#

The complete ionic reaction is when we separate out the soluble ions. Anything aqueous is soluble in water. Thus (and don't forget to distribute the coefficients!):

#3"Ag"^(+)(aq) + 3"NO"_3^(-)(aq) + 3"Na"^(+)(aq) + "PO"_4^(3-)(aq) -> "Ag"_3"PO"_4(s) + 3"Na"^(+)(aq) + 3"NO"_3^(-)(aq)#

Now, if we notice the spectator ions on each side, i.e. the ions that are common to each side, they don't do squat. So:

#3"Ag"^(+)(aq) + cancel(3"NO"_3^(-)(aq)) + cancel(3"Na"^(+)(aq)) + "PO"_4^(3-)(aq) -> "Ag"_3"PO"_4(s) + cancel(3"Na"^(+)(aq)) + cancel(3"NO"_3^(-)(aq))#

The net ionic reaction is the result of this cancellation of spectator ions:

#color(blue)(3"Ag"^(+)(aq) + "PO"_4^(3-)(aq) -> "Ag"_3"PO"_4(s))#

In essence, it is just a narrowed-down version of the important part of the reaction.