# Question 00a68

Sep 17, 2014

First, you must recognize that this is a double displacement reaction.

#### Explanation:

Next you must write the ionic formulas for the products.

You know that ${\text{NO}}_{3}$ has an ionic charge of -1 and $\text{S}$ has a charge of -2. So the ionic charge on $\text{B} i$ is +3, while on $\text{Ba}$ it is -2.

After the cations have changed partners, the formulas of the products are ${\text{Bi"_2"S}}_{3}$ 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.

• $\text{BaS}$ is soluble, because $\text{Ba}$ is in Group 2.

• ${\text{B"_2"S}}_{3}$ is insoluble, because $\text{Bi}$ is in group 15.

The unbalanced molecular equation is

$\text{Bi"("NO"_3)_3"(aq)" + "BaS (aq") → "Bi"_2"S"_3"(s)" + "Ba"("NO"_3)_2"(aq)}$

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

$\text{2Bi"^(3+)"(aq)" + "6NO"_3^(-)"(aq)" + "3Ba"^(2+)"(aq)" + "3S"^(2-)"(aq)" → "Bi"_2"S"_3"(s)" + "3Ba"^(2+)"(aq)"+ "6NO"_3^(-)"(aq)}$

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)")))#

$\text{2Bi"^(3+)"(aq)" + "3S"^(2-)"(aq)" → "Bi"_2"S"_3"(s)}$

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