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.