Question #6b77f

1 Answer
Jul 21, 2015


Here's how you can figure these out.


Once again, it is very useful to use state symbols to show which compound is ionic and which is covalent.

This will help ypu determine which compound will exist as ions in aqueous solution and which as molecules.

In your case, you have two soluble ionic compounds, sodium sulfite, #Na_2SO_3#, and barium acetate, #Ba(C_2H_3O_2)_2#, reacting to produce a soluble salt, sodium acetate, #NaC_2H_3O_2#, and an insoluble solid, barium sulfite, #BaSO_3#, which will precipitate out of solution.

This means that barium sulfite will not exist as ions in solution. So, let's take these equations one by one

  • Molecular equation

To write the molecular equation, simply use the compounds as they are written - but don't forget to include state symbols!

#Na_2SO_(3(aq)) + Ba(CH_3COO)_(2(aq)) -> color(green)(2)NaCH_3COO_((aq)) + BaSO_(3(s)) darr#

  • Complete ionic equation

To write the complete ionic equation you need to break all the ionic compounds into cations and anions. All ions that take part in the reaction are included.

#color(blue)(2)Na_((aq))^(+) + SO_(3(aq))^(2-) + Ba_((aq))^(2+) + color(red)(2)CH_3COO_((aq))^(-) -> color(green)(2)Na_((aq))^(+) + color(green)(2)CH_3COO_((aq))^(-) + BaSO_(3(s)) darr#

  • Net ionic equation

This time you need to remove spectator ions, which are ions that are present on both sides of the equation.

In your case, the sodium cations and the acetate anions will not appear in the net ionic equation.

#cancel(color(blue)(2)Na_((aq))^(+)) + SO_(3(aq))^(2-) + Ba_((aq))^(2+) + cancel(color(red)(2)CH_3COO_((aq))^(-)) -> cancel(color(green)(2)Na_((aq))^(+)) + cancel(color(green)(2)CH_3COO_((aq))^(-)) + BaSO_(3(s)) darr#

This is equivalent to

#SO_(3(aq))^(2-) + Ba_((aq))^(2+) -> BaSO_(3(s)) darr#