Question #00a68

Next you must write the ionic formulas for the products.

You know that `"NO"_3` has an ionic charge of -1 and `"S"` has a charge of -2. So the ionic charge on `"B"i` is +3, while on `"Ba"` it is -2.

After the cations have changed partners, the formulas of the products are `"Bi"_2"S"_3` and `"Ba"("NO"_3)_2`.

Next, you use the solubility rules to determine if there are any precipitates.

In this case, the important rules are:

1. All nitrates are soluble.
2. All sulfides are insoluble except those of Groups 1 and 2.

So

• `"Ba"("NO"_3)_2` is soluble.

• `"BaS"` is soluble, because `"Ba"` is in Group 2.

• `"B"_2"S"_3` is insoluble, because `"Bi"` is in group 15.

The unbalanced molecular equation is

`"Bi"("NO"_3)_3"(aq)" + "BaS (aq") → "Bi"_2"S"_3"(s)" + "Ba"("NO"_3)_2"(aq)"`

The balanced molecular equation is

`"2Bi"("NO"_3)_3"(aq)" + "3BaS(aq)" → "Bi"_2"S"_3"(s)" + "3Ba"("NO"_3)_2(aq)`

The total ionic equation is

`"2Bi"^(3+)"(aq)" + "6NO"_3^(-)"(aq)" + "3Ba"^(2+)"(aq)" + "3S"^(2-)"(aq)" → "Bi"_2"S"_3"(s)" + "3Ba"^(2+)"(aq)"+ "6NO"_3^(-)"(aq)"`

To get the net ionic equation, we cancel the spectator ions.

`"2Bi"^(3+)"(aq)" + color(red)(cancel(color(black)("6NO"_3^(-)"(aq)"))) + color(red)(cancel(color(black)("3Ba"^(2+)"(aq)"))) + "3S"^(2-)"(aq)" → "Bi"_2"S"_3"(s)" + color(red)(cancel(color(black)("3Ba"^(2+)"(aq)"))) + color(red)(cancel(color(black)("6NO"_3^(-)"(aq)")))`

`"2Bi"^(3+)"(aq)" + "3S"^(2-)"(aq)" → "Bi"_2"S"_3"(s)"`

Here's a good video on solubility rules and net ionic equations.